CBSE Class 10 Maths Chapter 2 Polynomials Model Questions - 5 Marks - Part 1
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CBSE Class 10 Maths Chapter 2 Polynomials Model Questions - 5 Marks - Part 1
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SECTION E — Long Answer Type Questions
[5 Marks Each]
Q1. If $\alpha$ and $\beta$ are the zeroes of the quadratic polynomial $f(x) = 3x^2 - 4x + 1$, evaluate the exact numerical value of the following expressions without directly computing the individual values of $\alpha$ and $\beta$:
Answer:
1. Identify the coefficients from the polynomial $f(x) = 3x^2 - 4x + 1$ in standard form $ax^2 + bx + c$:
$a = 3$, $b = -4$, $c = 1$.
2. Find the sum and product of the zeroes ($\alpha$ and $\beta$) using coefficient relationships:
• Sum ($\alpha + \beta$) = $-\frac{b}{a} = -\frac{-4}{3} = \frac{4}{3}$
• Product ($\alpha\beta$) = $\frac{c}{a} = \frac{1}{3}$
(a) Evaluate $\alpha^3 + \beta^3$:
1. Use the algebraic identity for the sum of cubes: $\alpha^3 + \beta^3 = (\alpha + \beta)^3 - 3\alpha\beta(\alpha + \beta)$
2. Substitute the values of the sum ($\frac{4}{3}$) and product ($\frac{1}{3}$) into the identity:
$\alpha^3 + \beta^3 = \left(\frac{4}{3}\right)^3 - 3\left(\frac{1}{3}\right)\left(\frac{4}{3}\right)$
$\alpha^3 + \beta^3 = \frac{64}{27} - \frac{4}{3}$
3. Convert $\frac{4}{3}$ to a common denominator of $27$:
$\frac{4}{3} = \frac{36}{27}$
4. Subtract the fractions:
$\alpha^3 + \beta^3 = \frac{64}{27} - \frac{36}{27} = \frac{28}{27}$
Result for (a): $\frac{28}{27}$
(b) Evaluate $\frac{\alpha^2}{\beta} + \frac{\beta^2}{\alpha}$:
1. Simplify the expression by finding a common denominator:
$\frac{\alpha^2}{\beta} + \frac{\beta^2}{\alpha} = \frac{\alpha^3 + \beta^3}{\alpha\beta}$
2. Substitute the numerator value found in part (a) ($\frac{28}{27}$) and the product ($\frac{1}{3}$):
$\frac{\frac{28}{27}}{\frac{1}{3}} = \frac{28}{27} \times \frac{3}{1} = \frac{84}{27}$
3. Simplify the fraction by dividing the numerator and denominator by their greatest common divisor ($3$):
$\frac{84 \div 3}{27 \div 3} = \frac{28}{9}$
Result for (b): $\frac{28}{9}$
Q2. Find the zeroes of the quadratic polynomial $p(x) = 4\sqrt{3}x^2 + 5x - 2\sqrt{3}$ by using the method of splitting the middle term. Verify the relationship between the zeroes and the coefficients rigorously by computing both the sum and product relationships.
Answer: Step 1: Find the zeroes by splitting the middle term
1. Set the polynomial equal to zero: $4\sqrt{3}x^2 + 5x - 2\sqrt{3} = 0$.
2. Multiply the leading coefficient ($a = 4\sqrt{3}$) by the constant term ($c = -2\sqrt{3}$):
$(4\sqrt{3}) \times (-2\sqrt{3}) = -8 \times 3 = -24$.
3. Find two numbers that multiply to $-24$ and add up to the middle coefficient ($5$). These numbers are $8$ and $-3$.
4. Split the middle term using $8x$ and $-3x$:
$4\sqrt{3}x^2 + 8x - 3x - 2\sqrt{3} = 0$
6. Solve for the zeroes ($\alpha$ and $\beta$):
• $4x - \sqrt{3} = 0 \implies x = \frac{\sqrt{3}}{4}$
• $\sqrt{3}x + 2 = 0 \implies x = -\frac{2}{\sqrt{3}}$
Let $\alpha = \frac{\sqrt{3}}{4}$ and $\beta = -\frac{2}{\sqrt{3}}$.
Step 2: Verify the relationship between zeroes and coefficients
From the polynomial $p(x) = 4\sqrt{3}x^2 + 5x - 2\sqrt{3}$, the coefficients are:
$a = 4\sqrt{3}$, $b = 5$, $c = -2\sqrt{3}$.
1. Verification of Sum of Zeroes:
• Actual Sum ($\alpha + \beta$) = $\frac{\sqrt{3}}{4} + \left(-\frac{2}{\sqrt{3}}\right) = \frac{3 - 8}{4\sqrt{3}} = \frac{-5}{4\sqrt{3}}$
• Formula Sum ($-\frac{b}{a}$) = $-\frac{5}{4\sqrt{3}}$
Since $\text{Actual Sum} = \text{Formula Sum}$, the relationship is verified.
2. Verification of Product of Zeroes:
• Actual Product ($\alpha\beta$) = $\left(\frac{\sqrt{3}}{4}\right) \times \left(-\frac{2}{\sqrt{3}}\right) = -\frac{2\sqrt{3}}{4\sqrt{3}} = -\frac{2}{4} = -\frac{1}{2}$
• Formula Product ($\frac{c}{a}$) = $\frac{-2\sqrt{3}}{4\sqrt{3}} = -\frac{2}{4} = -\frac{1}{2}$
Since $\text{Actual Product} = \text{Formula Product}$, the relationship is verified.
Result: Zeroes are $\frac{\sqrt{3}}{4}$ and $-\frac{2}{\sqrt{3}}$, and relationships are successfully verified.
Q3. If $\alpha$ and $\beta$ are the zeroes of the quadratic polynomial $g(x) = 2x^2 + 5x + k$, calculate the value of the unknown constant $k$ if it is given that the zeroes satisfy the symmetric condition:
Answer:
1. Identify the coefficients from the polynomial $g(x) = 2x^2 + 5x + k$ in standard form $ax^2 + bx + c$:
$a = 2$, $b = 5$, $c = k$.
2. Express the sum and product of the zeroes ($\alpha$ and $\beta$) in terms of these coefficients:
• Sum ($\alpha + \beta$) = $-\frac{b}{a} = -\frac{5}{2}$
• Product ($\alpha\beta$) = $\frac{c}{a} = \frac{k}{2}$
3. Rewrite the given condition $\alpha^2 + \beta^2 + \alpha\beta = \frac{21}{4}$ by expanding $\alpha^2 + \beta^2$ in terms of $(\alpha + \beta)^2$:
$\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta$
So the condition becomes:
$(\alpha + \beta)^2 - 2\alpha\beta + \alpha\beta = \frac{21}{4}$
$(\alpha + \beta)^2 - \alpha\beta = \frac{21}{4}$
4. Substitute the expressions for the sum and product into the equation:
$\left(-\frac{5}{2}\right)^2 - \frac{k}{2} = \frac{21}{4}$
Answer:
1. Identify the coefficients from the polynomial $f(x) = x^2 - px + q$ in standard form $ax^2 + bx + c$:
$a = 1$, $b = -p$, $c = q$.
2. Find the sum and product of the zeroes ($\alpha$ and $\beta$) using coefficient relationships:
• Sum ($\alpha + \beta$) = $-\frac{b}{a} = -\frac{-p}{1} = p$
• Product ($\alpha\beta$) = $\frac{c}{a} = \frac{q}{1} = q$
3. Simplify the expression to be proved by finding a common denominator:
$\frac{\alpha^2}{\beta^2} + \frac{\beta^2}{\alpha^2} = \frac{(\alpha^2)^2 + (\beta^2)^2}{(\alpha\beta)^2} = \frac{\alpha^4 + \beta^4}{(\alpha\beta)^2}$
6. Substitute the numerator back into the fraction with the denominator $(\alpha\beta)^2 = q^2$:
$\frac{\alpha^4 + \beta^4}{(\alpha\beta)^2} = \frac{p^4 - 4p^2q + 2q^2}{q^2}$
7. Split the fraction into separate terms:
$\frac{p^4}{q^2} - \frac{4p^2q}{q^2} + \frac{2q^2}{q^2} = \frac{p^4}{q^2} - \frac{4p^2}{q} + 2$
Q5. Find the zeroes of the quadratic polynomial $f(x) = 2x^2 - (1 + 2\sqrt{2})x + \sqrt{2}$. After determining the zeroes explicitly, establish and verify the relationship between its zeroes and its structural coefficients.
Answer: Step 1: Find the zeroes by splitting the middle term
1. Set the polynomial equal to zero: $2x^2 - (1 + 2\sqrt{2})x + \sqrt{2} = 0$.
2. Multiply the leading coefficient ($a = 2$) by the constant term ($c = \sqrt{2}$):
$2 \times \sqrt{2} = 2\sqrt{2}$.
3. The middle coefficient is $-(1 + 2\sqrt{2})$. We need two numbers that multiply to $2\sqrt{2}$ and add up to $-(1 + 2\sqrt{2})$. These numbers are $-1$ and $-2\sqrt{2}$.
4. Split the middle term:
$2x^2 - x - 2\sqrt{2}x + \sqrt{2} = 0$
6. Solve for the zeroes ($\alpha$ and $\beta$):
• $x - \sqrt{2} = 0 \implies x = \sqrt{2}$
• $2x - 1 = 0 \implies x = \frac{1}{2}$
Let $\alpha = \sqrt{2}$ and $\beta = \frac{1}{2}$.
Step 2: Verify the relationship between zeroes and coefficients
From the polynomial $f(x) = 2x^2 - (1 + 2\sqrt{2})x + \sqrt{2}$, the coefficients are:
$a = 2$, $b = -(1 + 2\sqrt{2})$, $c = \sqrt{2}$.
1. Verification of Sum of Zeroes:
• Actual Sum ($\alpha + \beta$) = $\sqrt{2} + \frac{1}{2} = \frac{2\sqrt{2} + 1}{2}$
• Formula Sum ($-\frac{b}{a}$) = $-\frac{-(1 + 2\sqrt{2})}{2} = \frac{1 + 2\sqrt{2}}{2}$
Since $\text{Actual Sum} = \text{Formula Sum}$, the relationship is verified.
2. Verification of Product of Zeroes:
• Actual Product ($\alpha\beta$) = $\sqrt{2} \times \frac{1}{2} = \frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}}$
• Formula Product ($\frac{c}{a}$) = $\frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}}$
Since $\text{Actual Product} = \text{Formula Product}$, the relationship is verified.
Result: Zeroes are $\sqrt{2}$ and $\frac{1}{2}$, and relationships are successfully verified.
Q6. If $\alpha$ and $\beta$ are the zeroes of the polynomial $p(x) = 3x^2 - 6x + 4$, evaluate the numerical value of the expression:
Answer:
1. Identify the coefficients from the polynomial $p(x) = 3x^2 - 6x + 4$ in standard form $ax^2 + bx + c$:
$a = 3$, $b = -6$, $c = 4$.
2. Find the sum and product of the zeroes ($\alpha$ and $\beta$) using coefficient relationships:
• Sum ($\alpha + \beta$) = $-\frac{b}{a} = -\frac{-6}{3} = \frac{6}{3} = 2$
• Product ($\alpha\beta$) = $\frac{c}{a} = \frac{4}{3}$
3. Simplify each part of the given expression in terms of $(\alpha + \beta)$ and $\alpha\beta$: Part 1: $\frac{\alpha}{\beta} + \frac{\beta}{\alpha} = \frac{\alpha^2 + \beta^2}{\alpha\beta} = \frac{(\alpha + \beta)^2 - 2\alpha\beta}{\alpha\beta}$
Substitute the sum ($2$) and product ($\frac{4}{3}$):
$\frac{(2)^2 - 2\left(\frac{4}{3}\right)}{\frac{4}{3}} = \frac{4 - \frac{8}{3}}{\frac{4}{3}} = \frac{\frac{4}{3}}{\frac{4}{3}} = 1$
Part 2: $2\left(\frac{1}{\alpha} + \frac{1}{\beta}\right) = 2\left(\frac{\alpha + \beta}{\alpha\beta}\right)$
Substitute the sum ($2$) and product ($\frac{4}{3}$):
$2\left(\frac{2}{\frac{4}{3}}\right) = 2\left(2 \times \frac{3}{4}\right) = 2\left(\frac{6}{4}\right) = 2\left(\frac{3}{2}\right) = 3$
Part 3: $3\alpha\beta$
Substitute the product ($\frac{4}{3}$):
$3\left(\frac{4}{3}\right) = 4$
4. Combine all evaluated parts together:
$1 + 3 + 4 = 8$
Result: $8$
Q7. If $\alpha$ and $\beta$ are the zeroes of the quadratic polynomial $f(x) = 6x^2 - 5x + 1$, find the value of:
$$\frac{1}{\alpha^3} + \frac{1}{\beta^3}$$
Answer:
1. Identify the coefficients from the polynomial $f(x) = 6x^2 - 5x + 1$ in standard form $ax^2 + bx + c$:
$a = 6$, $b = -5$, $c = 1$.
2. Find the sum and product of the zeroes ($\alpha$ and $\beta$) using coefficient relationships:
• Sum ($\alpha + \beta$) = $-\frac{b}{a} = -\frac{-5}{6} = \frac{5}{6}$
• Product ($\alpha\beta$) = $\frac{c}{a} = \frac{1}{6}$
3. Simplify the expression by finding a common denominator:
$\frac{1}{\alpha^3} + \frac{1}{\beta^3} = \frac{\alpha^3 + \beta^3}{(\alpha\beta)^3}$
5. Substitute the sum ($\frac{5}{6}$) and product ($\frac{1}{6}$) into the numerator expression:
$\alpha^3 + \beta^3 = \left(\frac{5}{6}\right)^3 - 3\left(\frac{1}{6}\right)\left(\frac{5}{6}\right)$
$\alpha^3 + \beta^3 = \frac{125}{216} - \frac{15}{36}$
6. Convert $\frac{15}{36}$ to a denominator of $216$ (since $36 \times 6 = 216$ and $15 \times 6 = 90$):
$\frac{125}{216} - \frac{90}{216} = \frac{35}{216}$
7. Substitute the numerator and denominator $(\alpha\beta)^3 = \left(\frac{1}{6}\right)^3 = \frac{1}{216}$ back into the fraction:
$\frac{\frac{35}{216}}{\frac{1}{216}} = \frac{35}{216} \times \frac{216}{1} = 35$
Result: $35$
Q8. If $\alpha$ and $\beta$ are the zeroes of the quadratic polynomial $f(x) = x^2 - kx + 6$ such that the difference between their squares is equal to 13 (i.e., $\alpha^2 - \beta^2 = 13$), find the possible values of the parameter $k$.
Answer:
1. Identify the coefficients from the polynomial $f(x) = x^2 - kx + 6$ in standard form $ax^2 + bx + c$:
$a = 1$, $b = -k$, $c = 6$.
2. Express the sum and product of the zeroes ($\alpha$ and $\beta$) using coefficient relationships:
• Sum ($\alpha + \beta$) = $-\frac{b}{a} = -\frac{-k}{1} = k$
• Product ($\alpha\beta$) = $\frac{c}{a} = \frac{6}{1} = 6$
3. Factor the given condition $\alpha^2 - \beta^2 = 13$ using the difference of squares identity:
$(\alpha - \beta)(\alpha + \beta) = 13$
5. Substitute $(\alpha + \beta) = k$ and $(\alpha - \beta) = \pm\sqrt{k^2 - 24}$ into the factored condition:
$\pm\sqrt{k^2 - 24} \cdot k = 13$
6. Square both sides to eliminate the square root:
$(k^2 - 24)k^2 = 13^2$
$k^4 - 24k^2 = 169$
$k^4 - 24k^2 - 169 = 0$
7. Let $y = k^2$, turning the equation into a quadratic in terms of $y$:
$y^2 - 24y - 169 = 0$
8. Solve for $y$ using the quadratic formula $y = \frac{-B \pm \sqrt{B^2 - 4AC}}{2A}$ where $A = 1, B = -24, C = -169$:
$y = \frac{-(-24) \pm \sqrt{(-24)^2 - 4(1)(-169)}}{2(1)}$
$y = \frac{24 \pm \sqrt{576 + 676}}{2} = \frac{24 \pm \sqrt{1252}}{2}$
Simplify $\sqrt{1252} = \sqrt{4 \times 313} = 2\sqrt{313}$:
$y = \frac{24 \pm 2\sqrt{313}}{2} = 12 \pm \sqrt{313}$
9. Since $y = k^2$, and $12 - \sqrt{313}$ is negative (since $\sqrt{313} > \sqrt{289} = 17$), we take the positive value:
$k^2 = 12 + \sqrt{313}$
$k = \pm\sqrt{12 + \sqrt{313}}$
Result: $k = \pm\sqrt{12 + \sqrt{313}}$
Q9. If one zero of the quadratic polynomial $p(x) = (k^2 + 4)x^2 + 13x + 4k$ is the reciprocal of the other zero, find the numerical value of $k$. Using this value of $k$, rewrite the polynomial and calculate its actual zeroes.
Answer: Step 1: Find the value of $k$
1. Identify the coefficients from the polynomial $p(x) = (k^2 + 4)x^2 + 13x + 4k$ in standard form $ax^2 + bx + c$:
$a = k^2 + 4$, $b = 13$, $c = 4k$.
2. Let one zero be $\alpha$. Since the other zero is its reciprocal, it is $\frac{1}{\alpha}$.
3. Use the product of the zeroes relationship: $\text{Product} = \alpha \times \frac{1}{\alpha} = 1$.
4. Alternatively, express the product of the zeroes using coefficients ($\frac{c}{a}$):
$\text{Product} = \frac{4k}{k^2 + 4}$
5. Equate the product to $1$ since one zero is the reciprocal of the other:
$\frac{4k}{k^2 + 4} = 1$
6. Solve for $k$ by cross-multiplication:
$4k = k^2 + 4$
$k^2 - 4k + 4 = 0$
7. Factor the quadratic equation:
$(k - 2)^2 = 0 \implies k = 2$
Value of $k$: $k = 2$
Step 2: Rewrite the polynomial and find its actual zeroes
1. Substitute $k = 2$ back into the polynomial $p(x)$:
$p(x) = (2^2 + 4)x^2 + 13x + 4(2)$
$p(x) = (4 + 4)x^2 + 13x + 8$
$p(x) = 8x^2 + 13x + 8$
2. Find the zeroes of $p(x) = 8x^2 + 13x + 8$ using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ where $a = 8, b = 13, c = 8$:
$x = \frac{-13 \pm \sqrt{(13)^2 - 4(8)(8)}}{2(8)}$
$x = \frac{-13 \pm \sqrt{169 - 256}}{16}$
$x = \frac{-13 \pm \sqrt{-87}}{16}$
3. Since the discriminant is negative ($\Delta = -87$), the zeroes are complex numbers:
$x = \frac{-13 \pm i\sqrt{87}}{16}$
Result: $k = 2$, the rewritten polynomial is $p(x) = 8x^2 + 13x + 8$, and its zeroes are $\frac{-13 \pm i\sqrt{87}}{16}$.
Q10. If $\alpha$ and $\beta$ are the zeroes of the polynomial $f(x) = 3x^2 - 5x - 2$, find the value of $k$ if it is given that $2\alpha + 3\beta = k$ and $\alpha - \beta = \frac{7}{3}$.
Answer:
1. Identify the coefficients from the polynomial $f(x) = 3x^2 - 5x - 2$ in standard form $ax^2 + bx + c$:
$a = 3$, $b = -5$, $c = -2$.
2. Find the sum and product of the zeroes ($\alpha$ and $\beta$) using coefficient relationships:
• Sum ($\alpha + \beta$) = $-\frac{b}{a} = -\frac{-5}{3} = \frac{5}{3}$
• Product ($\alpha\beta$) = $\frac{c}{a} = \frac{-2}{3}$
3. We are given two equations involving $\alpha$ and $\beta$:
① $\alpha + \beta = \frac{5}{3}$
② $\alpha - \beta = \frac{7}{3}$
4. Solve this system of linear equations to find the exact values of $\alpha$ and $\beta$:
Add equation ① and equation ②:
$(\alpha + \beta) + (\alpha - \beta) = \frac{5}{3} + \frac{7}{3}$
$2\alpha = \frac{12}{3} = 4 \implies \alpha = 2$
5. Substitute $\alpha = 2$ into equation ① to find $\beta$:
$2 + \beta = \frac{5}{3}$
$\beta = \frac{5}{3} - 2 = \frac{5 - 6}{3} = -\frac{1}{3}$
6. Verify if these zeroes satisfy the product condition ($\alpha\beta = -\frac{2}{3}$):
$2 \times \left(-\frac{1}{3}\right) = -\frac{2}{3}$ (Verified)
7. Use the given condition for $k$: $2\alpha + 3\beta = k$
Substitute $\alpha = 2$ and $\beta = -\frac{1}{3}$ into the equation:
$k = 2(2) + 3\left(-\frac{1}{3}\right)$
$k = 4 - 1 = 3$
Result: $k = 3$
Q11. If the sum of the zeroes of the quadratic polynomial $p(t) = kt^2 + 3t + 4k$ is exactly equal to twice the product of their zeroes, determine the value of $k$. Hence, find the sum and the product of the zeroes of this modified polynomial.
Answer: Step 1: Determine the value of $k$
1. Identify the coefficients from the polynomial $p(t) = kt^2 + 3t + 4k$ in standard form $at^2 + bt + c$:
$a = k$, $b = 3$, $c = 4k$.
2. Express the sum and product of the zeroes using coefficient relationships:
• Sum = $-\frac{b}{a} = -\frac{3}{k}$
• Product = $\frac{c}{a} = \frac{4k}{k} = 4$
3. Use the given condition: "sum of the zeroes is equal to twice the product of their zeroes":
$\text{Sum} = 2 \times (\text{Product})$
$-\frac{3}{k} = 2(4)$
4. Solve for $k$:
$-\frac{3}{k} = 8$
$8k = -3 \implies k = -\frac{3}{8}$
Value of $k$: $k = -\frac{3}{8}$
Step 2: Find the sum and product of the zeroes for the modified polynomial
1. Substitute $k = -\frac{3}{8}$ into the expressions for the sum and product:
• Sum = $-\frac{3}{-\frac{3}{8}} = -3 \times \left(-\frac{8}{3}\right) = 8$
• Product = $4$ (Notice that the product $\frac{4k}{k} = 4$ is independent of $k$ as long as $k \neq 0$)
2. Alternatively, rewrite the polynomial with $k = -\frac{3}{8}$:
$p(t) = -\frac{3}{8}t^2 + 3t + 4\left(-\frac{3}{8}\right) = -\frac{3}{8}t^2 + 3t - \frac{3}{2}$
Using coefficients of the rewritten polynomial ($a = -\frac{3}{8}, b = 3, c = -\frac{3}{2}$):
• Sum = $-\frac{3}{-\frac{3}{8}} = 8$
• Product = $\frac{-\frac{3}{2}}{-\frac{3}{8}} = \frac{3}{2} \times \frac{8}{3} = 4$
Result: $k = -\frac{3}{8}$, the sum of the zeroes is $8$, and the product of the zeroes is $4$.
Q12. If $\alpha$ and $\beta$ are the zeroes of the polynomial $f(x) = x^2 - 6x + k$, find the value of $k$ if it satisfies the linear combination equation $3\alpha + 2\beta = 20$.
Answer:
1. Identify the coefficients from the polynomial $f(x) = x^2 - 6x + k$ in standard form $ax^2 + bx + c$:
$a = 1$, $b = -6$, $c = k$.
2. Find the sum and product of the zeroes ($\alpha$ and $\beta$) using coefficient relationships:
• Sum ($\alpha + \beta$) = $-\frac{b}{a} = -\frac{-6}{1} = 6$
• Product ($\alpha\beta$) = $\frac{c}{a} = \frac{k}{1} = k$
3. We are given two equations involving $\alpha$ and $\beta$:
① $\alpha + \beta = 6$
② $3\alpha + 2\beta = 20$
4. Solve this system of linear equations to find the exact values of $\alpha$ and $\beta$:
Multiply equation ① by $2$:
$2\alpha + 2\beta = 12$ — (Equation ③)
Subtract Equation ③ from Equation ②:
$(3\alpha + 2\beta) - (2\alpha + 2\beta) = 20 - 12$
$\alpha = 8$
5. Substitute $\alpha = 8$ into equation ① to find $\beta$:
$8 + \beta = 6$
$\beta = 6 - 8 = -2$
6. Use the product relationship to find $k$ ($\alpha\beta = k$):
$k = 8 \times (-2) = -16$
Result: $k = -16$
Q13. If one zero of the polynomial $2x^2 - 3x + p$ is 3, find the value of $p$. Using this value of $p$, determine the second zero of the polynomial and verify the product of the zeroes against the constant term ratio.
Answer: Step 1: Find the value of $p$
1. Let the given polynomial be $f(x) = 2x^2 - 3x + p$.
2. Since $x = 3$ is a zero of the polynomial, substituting $x = 3$ must result in $f(3) = 0$:
$2(3)^2 - 3(3) + p = 0$
3. Simplify the equation:
$2(9) - 9 + p = 0$
$18 - 9 + p = 0$
$9 + p = 0 \implies p = -9$
Value of $p$: $p = -9$
Step 2: Determine the second zero
1. Substitute $p = -9$ back into the polynomial:
$f(x) = 2x^2 - 3x - 9$
2. Factor the polynomial by splitting the middle term (product $2 \times (-9) = -18$, sum $-3$):
$2x^2 - 6x + 3x - 9 = 0$
$2x(x - 3) + 3(x - 3) = 0$
$(2x + 3)(x - 3) = 0$
3. Find the zeroes:
• $x - 3 = 0 \implies x = 3$ (given zero)
• $2x + 3 = 0 \implies x = -\frac{3}{2}$
Second zero: $-\frac{3}{2}$
Step 3: Verify the product of the zeroes
1. Calculate the actual product of the zeroes ($\alpha$ and $\beta$):
$\text{Actual Product} = 3 \times \left(-\frac{3}{2}\right) = -\frac{9}{2}$
2. Calculate the formula product ($\frac{c}{a}$ from $2x^2 - 3x - 9$ where $a = 2, c = -9$):
$\text{Formula Product} = \frac{-9}{2}$
3. Since $\text{Actual Product} = \text{Formula Product} = -\frac{9}{2}$, the relationship is verified.
Result: $p = -9$, the second zero is $-\frac{3}{2}$, and the product of the zeroes is successfully verified.
Q14. If the product of the zeroes of the quadratic polynomial $g(x) = ax^2 - 6x - 6$ is equal to 4, determine the value of the leading coefficient $a$. With this value of $a$, verify the relation for the sum of the zeroes.
Answer: Step 1: Determine the value of $a$
1. Identify the coefficients from the polynomial $g(x) = ax^2 - 6x - 6$ in standard form:
$a = a$, $b = -6$, $c = -6$.
2. Express the product of the zeroes using coefficient relationships ($\frac{c}{a}$):
$\text{Product} = \frac{-6}{a}$
3. Use the given condition that the product of the zeroes is equal to $4$:
$\frac{-6}{a} = 4$
Step 2: Verify the relation for the sum of the zeroes
1. Rewrite the polynomial with $a = -\frac{3}{2}$:
$g(x) = -\frac{3}{2}x^2 - 6x - 6$
2. Find the zeroes of the polynomial by setting $g(x) = 0$:
$-\frac{3}{2}x^2 - 6x - 6 = 0$
Multiply the entire equation by $-\frac{2}{3}$ to simplify:
$x^2 + 4x + 4 = 0$
$(x + 2)^2 = 0 \implies x = -2, -2$
Thus, the zeroes are $\alpha = -2$ and $\beta = -2$.
3. Verification of Sum of Zeroes:
• Actual Sum ($\alpha + \beta$) = $(-2) + (-2) = -4$
• Formula Sum ($-\frac{b}{a}$) = $-\frac{-6}{-\frac{3}{2}} = -\left(-6 \times \frac{2}{3}\right) = -(-4) = -4$
Since $\text{Actual Sum} = \text{Formula Sum} = -4$, the relationship is verified.
Result: $a = -\frac{3}{2}$, the zeroes are $-2$ and $-2$, and the sum of the zeroes is successfully verified.
Q15. If $\alpha$ and $\beta$ are the zeroes of the quadratic polynomial $f(x) = 2x^2 - 5x + 3$, construct a brand new quadratic polynomial whose zeroes are defined as $\frac{1}{\alpha}$ and $\frac{1}{\beta}$.
Answer: Step 1: Find the sum and product of the original zeroes
1. Identify the coefficients from the polynomial $f(x) = 2x^2 - 5x + 3$ in standard form $ax^2 + bx + c$:
$a = 2$, $b = -5$, $c = 3$.
2. Find the sum and product of the zeroes ($\alpha$ and $\beta$) using coefficient relationships:
• Sum ($\alpha + \beta$) = $-\frac{b}{a} = -\frac{-5}{2} = \frac{5}{2}$
• Product ($\alpha\beta$) = $\frac{c}{a} = \frac{3}{2}$
Step 2: Find the sum and product of the new zeroes
1. Let the new zeroes be $S_1 = \frac{1}{\alpha}$ and $S_2 = \frac{1}{\beta}$.
2. Calculate the sum of the new zeroes:
$\text{New Sum} = \frac{1}{\alpha} + \frac{1}{\beta} = \frac{\alpha + \beta}{\alpha\beta}$
Substitute the original sum ($\frac{5}{2}$) and product ($\frac{3}{2}$):
$\text{New Sum} = \frac{\frac{5}{2}}{\frac{3}{2}} = \frac{5}{2} \times \frac{2}{3} = \frac{5}{3}$
3. Calculate the product of the new zeroes:
$\text{New Product} = \left(\frac{1}{\alpha}\right) \times \left(\frac{1}{\beta}\right) = \frac{1}{\alpha\beta}$
Substitute the original product ($\frac{3}{2}$):
$\text{New Product} = \frac{1}{\frac{3}{2}} = \frac{2}{3}$
Step 3: Construct the new quadratic polynomial
1. A quadratic polynomial given its sum ($S$) and product ($P$) of zeroes is expressed as:
$p(x) = k \left(x^2 - (\text{New Sum})x + \text{New Product}\right)$ where $k$ is a non-zero real constant.
2. Substitute the new sum and product into the equation:
$p(x) = k \left(x^2 - \frac{5}{3}x + \frac{2}{3}\right)$
3. Choosing $k = 3$ to clear the fractions gives the standard integer-coefficient polynomial:
$p(x) = 3x^2 - 5x + 2$
Result: The new quadratic polynomial is $3x^2 - 5x + 2$ (or any non-zero scalar multiple thereof).
Q16. If $\alpha$ and $\beta$ are the zeroes of the polynomial $g(x) = x^2 - 3x - 2$, form a quadratic polynomial whose zeroes are:
Answer: Step 1: Find the sum and product of the original zeroes
1. Identify the coefficients from the polynomial $g(x) = x^2 - 3x - 2$ in standard form $ax^2 + bx + c$:
$a = 1$, $b = -3$, $c = -2$.
2. Find the sum and product of the zeroes ($\alpha$ and $\beta$) using coefficient relationships:
• Sum ($\alpha + \beta$) = $-\frac{b}{a} = -\frac{-3}{1} = 3$
• Product ($\alpha\beta$) = $\frac{c}{a} = \frac{-2}{1} = -2$
Step 2: Find the sum and product of the new zeroes
1. Let the new zeroes be $S_1 = 2\alpha + 3\beta$ and $S_2 = 3\alpha + 2\beta$.
2. Calculate the sum of the new zeroes:
$\text{New Sum} = (2\alpha + 3\beta) + (3\alpha + 2\beta) = 5\alpha + 5\beta = 5(\alpha + \beta)$
Substitute the original sum ($3$):
$\text{New Sum} = 5(3) = 15$
3. Calculate the product of the new zeroes:
$\text{New Product} = (2\alpha + 3\beta)(3\alpha + 2\beta)$
Expand the expression:
$= 6\alpha^2 + 4\alpha\beta + 9\alpha\beta + 6\beta^2$
$= 6(\alpha^2 + \beta^2) + 13\alpha\beta$
Rewrite $\alpha^2 + \beta^2$ as $(\alpha + \beta)^2 - 2\alpha\beta$:
$= 6\left[(\alpha + \beta)^2 - 2\alpha\beta\right] + 13\alpha\beta$
$= 6(\alpha + \beta)^2 - 12\alpha\beta + 13\alpha\beta$
$= 6(\alpha + \beta)^2 + \alpha\beta$
Substitute the original sum ($3$) and product ($-2$):
$\text{New Product} = 6(3)^2 + (-2) = 6(9) - 2 = 54 - 2 = 52$
Step 3: Construct the new quadratic polynomial
1. A quadratic polynomial given its sum ($S$) and product ($P$) of zeroes is expressed as:
$p(x) = k \left(x^2 - (\text{New Sum})x + \text{New Product}\right)$ where $k$ is a non-zero real constant.
2. Substitute the new sum ($15$) and product ($52$) into the equation:
$p(x) = k(x^2 - 15x + 52)$
Result: The new quadratic polynomial is $x^2 - 15x + 52$ (or any non-zero scalar multiple thereof).
Q17. If $\alpha$ and $\beta$ are the zeroes of the quadratic polynomial $p(x) = 3x^2 - 4x + 1$, form a new quadratic polynomial whose zeroes are $\frac{\alpha^2}{\beta}$ and $\frac{\beta^2}{\alpha}$.
Answer: Step 1: Find the sum and product of the original zeroes
1. Identify the coefficients from the polynomial $p(x) = 3x^2 - 4x + 1$ in standard form $ax^2 + bx + c$:
$a = 3$, $b = -4$, $c = 1$.
2. Find the sum and product of the zeroes ($\alpha$ and $\beta$) using coefficient relationships:
• Sum ($\alpha + \beta$) = $-\frac{b}{a} = -\frac{-4}{3} = \frac{4}{3}$
• Product ($\alpha\beta$) = $\frac{c}{a} = \frac{1}{3}$
Step 2: Find the sum and product of the new zeroes
1. Let the new zeroes be $S_1 = \frac{\alpha^2}{\beta}$ and $S_2 = \frac{\beta^2}{\alpha}$.
2. Calculate the sum of the new zeroes:
$\text{New Sum} = \frac{\alpha^2}{\beta} + \frac{\beta^2}{\alpha} = \frac{\alpha^3 + \beta^3}{\alpha\beta}$
Using the formula $\alpha^3 + \beta^3 = (\alpha + \beta)^3 - 3\alpha\beta(\alpha + \beta)$ and substituting the values:
$\alpha^3 + \beta^3 = \left(\frac{4}{3}\right)^3 - 3\left(\frac{1}{3}\right)\left(\frac{4}{3}\right) = \frac{64}{27} - \frac{4}{3} = \frac{64}{27} - \frac{36}{27} = \frac{28}{27}$
Now, divide by the product ($\alpha\beta = \frac{1}{3}$):
$\text{New Sum} = \frac{\frac{28}{27}}{\frac{1}{3}} = \frac{28}{27} \times \frac{3}{1} = \frac{84}{27} = \frac{28}{9}$
3. Calculate the product of the new zeroes:
$\text{New Product} = \left(\frac{\alpha^2}{\beta}\right) \times \left(\frac{\beta^2}{\alpha}\right) = \frac{\alpha^2\beta^2}{\alpha\beta} = \alpha\beta$
Substitute the original product ($\frac{1}{3}$):
$\text{New Product} = \frac{1}{3}$
Step 3: Construct the new quadratic polynomial
1. A quadratic polynomial given its sum ($S$) and product ($P$) of zeroes is expressed as:
$p(x) = k \left(x^2 - (\text{New Sum})x + \text{New Product}\right)$ where $k$ is a non-zero real constant.
2. Substitute the new sum ($\frac{28}{9}$) and product ($\frac{1}{3}$) into the equation:
$p(x) = k \left(x^2 - \frac{28}{9}x + \frac{1}{3}\right)$
3. Choosing $k = 9$ to clear the fractions gives the standard integer-coefficient polynomial:
$p(x) = 9x^2 - 28x + 3$
Result: The new quadratic polynomial is $9x^2 - 28x + 3$ (or any non-zero scalar multiple thereof).
Q18. If $\alpha$ and $\beta$ are the zeroes of the polynomial $f(x) = x^2 - p(x + 1) - c$ such that $(\alpha + 1)(\beta + 1) = 1 - c$, prove this relation. Furthermore, form a quadratic polynomial whose zeroes are $\frac{1}{\alpha + 1}$ and $\frac{1}{\beta + 1}$.
Answer: Step 1: Simplify the polynomial and find coefficients
1. Rewrite the given polynomial $f(x) = x^2 - p(x + 1) - c$ in standard quadratic form $ax^2 + bx + c$:
$f(x) = x^2 - px - p - c$
$f(x) = x^2 - px - (p + c)$
3. Find the sum and product of the zeroes ($\alpha$ and $\beta$) using coefficient relationships:
• Sum ($\alpha + \beta$) = $-\frac{b}{a} = -\frac{-p}{1} = p$
• Product ($\alpha\beta$) = $\frac{c_{\text{poly}}}{a} = \frac{-(p + c)}{1} = -(p + c) = -p - c$
Step 2: Prove the given relation $(\alpha + 1)(\beta + 1) = 1 - c$
1. Expand the Left Hand Side (LHS):
$(\alpha + 1)(\beta + 1) = \alpha\beta + \alpha + \beta + 1$
2. Substitute the values of the sum ($\alpha + \beta = p$) and product ($\alpha\beta = -p - c$):
$= (-p - c) + p + 1$
$= -p - c + p + 1$
$= 1 - c$
3. Since $\text{LHS} = \text{RHS}$, the relation $(\alpha + 1)(\beta + 1) = 1 - c$ is successfully proved.
Step 3: Form a quadratic polynomial with new zeroes
1. Let the new zeroes be $S_1 = \frac{1}{\alpha + 1}$ and $S_2 = \frac{1}{\beta + 1}$.
2. Calculate the sum of the new zeroes:
$\text{New Sum} = \frac{1}{\alpha + 1} + \frac{1}{\beta + 1} = \frac{(\beta + 1) + (\alpha + 1)}{(\alpha + 1)(\beta + 1)}$
Substitute numerator sum $[(\alpha + \beta) + 2 = p + 2]$ and denominator $[1 - c]$:
$\text{New Sum} = \frac{p + 2}{1 - c}$
3. Calculate the product of the new zeroes:
$\text{New Product} = \left(\frac{1}{\alpha + 1}\right) \times \left(\frac{1}{\beta + 1}\right) = \frac{1}{(\alpha + 1)(\beta + 1)}$
Substitute the denominator $[1 - c]$:
$\text{New Product} = \frac{1}{1 - c}$
4. Construct the new quadratic polynomial using the sum ($S$) and product ($P$):
$p(x) = k \left(x^2 - (\text{New Sum})x + \text{New Product}\right)$ where $k$ is a non-zero real constant.
$p(x) = k \left(x^2 - \left(\frac{p + 2}{1 - c}\right)x + \frac{1}{1 - c}\right)$
Result: The relation is proved, and the new quadratic polynomial is $(1 - c)x^2 - (p + 2)x + 1$ (or any non-zero scalar multiple thereof).
Q19. If $\alpha$ and $\beta$ are the zeroes of the polynomial $h(x) = x^2 - 2x - 8$, form a new quadratic polynomial whose zeroes are $\alpha^2$ and $\beta^2$. Verify your results by finding the numerical values of the roots directly.
Answer: Step 1: Form the polynomial using sum and product of original zeroes
1. Identify the coefficients from the polynomial $h(x) = x^2 - 2x - 8$ in standard form $ax^2 + bx + c$:
$a = 1$, $b = -2$, $c = -8$.
2. Find the sum and product of the zeroes ($\alpha$ and $\beta$) using coefficient relationships:
• Sum ($\alpha + \beta$) = $-\frac{b}{a} = -\frac{-2}{1} = 2$
• Product ($\alpha\beta$) = $\frac{c}{a} = \frac{-8}{1} = -8$
3. Let the new zeroes be $S_1 = \alpha^2$ and $S_2 = \beta^2$.
4. Calculate the sum of the new zeroes:
$\text{New Sum} = \alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta$
Substitute the original sum ($2$) and product ($-8$):
$\text{New Sum} = (2)^2 - 2(-8) = 4 + 16 = 20$
5. Calculate the product of the new zeroes:
$\text{New Product} = \alpha^2\beta^2 = (\alpha\beta)^2$
Substitute the original product ($-8$):
$\text{New Product} = (-8)^2 = 64$
6. Construct the new quadratic polynomial:
$p(x) = k \left(x^2 - (\text{New Sum})x + \text{New Product}\right)$ where $k$ is a non-zero real constant.
$p(x) = x^2 - 20x + 64$ (taking $k = 1$).
Step 2: Verify by finding the numerical values of the roots directly
1. Factor the original polynomial $h(x) = x^2 - 2x - 8$ directly:
$x^2 - 4x + 2x - 8 = 0$
$x(x - 4) + 2(x - 4) = 0$
$(x + 2)(x - 4) = 0$
Thus, the original zeroes are $\alpha = 4$ and $\beta = -2$ (or vice versa).
2. Calculate the direct new zeroes ($\alpha^2$ and $\beta^2$):
• $\alpha^2 = 4^2 = 16$
• $\beta^2 = (-2)^2 = 4$
3. Form the polynomial directly using these numerical roots ($16$ and $4$):
$p(x) = (x - 16)(x - 4)$
$p(x) = x^2 - 4x - 16x + 64$
$p(x) = x^2 - 20x + 64$
Result: The new quadratic polynomial is $x^2 - 20x + 64$, which matches both methods successfully.
Q20. If one zero of a quadratic polynomial $p(x) = ax^2 + bx + c$ with rational coefficients is $3 + \sqrt{5}$, deduce the other zero. Form the quadratic polynomial assuming $a = 1$, and evaluate the value of $a^2 + b^2 + c^2$.
Answer: Step 1: Deduce the other zero
1. Since the quadratic polynomial has rational coefficients, irrational zeroes must occur in conjugate pairs.
2. Given that one zero is $\alpha = 3 + \sqrt{5}$, the other zero ($\beta$) must be its conjugate:
$\beta = 3 - \sqrt{5}$
Step 2: Form the quadratic polynomial assuming $a = 1$
1. Find the sum of the zeroes ($\alpha + \beta$):
$\alpha + \beta = (3 + \sqrt{5}) + (3 - \sqrt{5}) = 6$
2. Find the product of the zeroes ($\alpha\beta$):
$\alpha\beta = (3 + \sqrt{5})(3 - \sqrt{5}) = 3^2 - (\sqrt{5})^2 = 9 - 5 = 4$
3. A quadratic polynomial is given by $p(x) = a \left(x^2 - (\text{Sum})x + \text{Product}\right)$:
$p(x) = a(x^2 - 6x + 4)$
4. Given that $a = 1$, the polynomial becomes:
$p(x) = x^2 - 6x + 4$
Step 3: Evaluate the value of $a^2 + b^2 + c^2$
1. Identify the coefficients from the polynomial $p(x) = 1x^2 - 6x + 4$ in standard form $ax^2 + bx + c$:
$a = 1$, $b = -6$, $c = 4$.
Result: The other zero is $3 - \sqrt{5}$, the polynomial with $a = 1$ is $x^2 - 6x + 4$, and the value of $a^2 + b^2 + c^2$ is $53$.
Academic Repository Disclaimer & எச்சரிக்கை அறிவிப்பு :
English: These model solutions are structured to global academic standards for self-assessment and practice purposes. Students are advised to verify with official textbooks for board examinations.
தமிழ்: இந்த மாதிரி கணிதத் தீர்வுகள் மாணவர்களின் சுய கற்றல் மற்றும் பயிற்சித் திறனுக்காக உலகளாவிய கல்வித் தரத்தில் உருவாக்கப்பட்டுள்ளன. தேர்வுகளுக்குத் தயாராகும் போது உங்கள் அதிகாரப்பூர்வ பாடப்புத்தகத்துடன் சரிபார்க்கவும்.
