CBSE Class 10 Maths Chapter 2 Polynomials Model Questions - 3 Marks - Part 1
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CBSE Class 10 Maths Chapter 2 Polynomials Model Questions - 3 Marks - Part 1
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SECTION C — Short Answer Type Questions
[3 Marks Each]
Q1. Find the zeroes of the quadratic polynomial $f(x) = 6x^2 - 3 - 7x$ and verify the relationship between the zeroes and the coefficients.
Answer: Zeroes of the polynomial: $-\frac{1}{3}$ and $\frac{3}{2}$
Justification & Calculation:
First, rewrite the given polynomial in standard form $ax^2 + bx + c$:
$$f(x) = 6x^2 - 7x - 3$$
Comparing with the standard quadratic form, we get:
$$a = 6, \quad b = -7, \quad c = -3$$
Step 1: Find the zeroes by factorisation
We need to split the middle term $-7x$ into two parts whose sum is $-7$ and product is $6 \times (-3) = -18$. These numbers are $-9$ and $2$.
$$f(x) = 6x^2 - 9x + 2x - 3$$
$$f(x) = 3x(2x - 3) + 1(2x - 3)$$
$$f(x) = (3x + 1)(2x - 3)$$
To find the zeroes, set $f(x) = 0$:
$$(3x + 1)(2x - 3) = 0$$
$$3x + 1 = 0 \implies x = -\frac{1}{3}$$
$$2x - 3 = 0 \implies x = \frac{3}{2}$$
Let $\alpha = -\frac{1}{3}$ and $\beta = \frac{3}{2}$.
Q2. Find the zeroes of the polynomial $g(x) = 4\sqrt{3}x^2 + 5x - 2\sqrt{3}$ by factorization and verify the relationship between its zeroes and coefficients.
Answer: Zeroes of the polynomial: $-\frac{2}{\sqrt{3}}$ (or $-\frac{2\sqrt{3}}{3}$) and $\frac{\sqrt{3}}{4}$
Justification & Calculation:
Given polynomial: $g(x) = 4\sqrt{3}x^2 + 5x - 2\sqrt{3}$
Comparing this with the standard quadratic form $ax^2 + bx + c$, we get:
$$a = 4\sqrt{3}, \quad b = 5, \quad c = -2\sqrt{3}$$
Step 1: Find the zeroes by factorisation
We need to split the middle term $5x$ into two parts whose sum is $5$ and whose product is equal to $a \cdot c$:
$$a \cdot c = (4\sqrt{3}) \cdot (-2\sqrt{3}) = -8 \cdot 3 = -24$$
The two numbers whose product is $-24$ and whose sum is $5$ are $8$ and $-3$.
Rewrite the middle term using $8x$ and $-3x$:
$$g(x) = 4\sqrt{3}x^2 + 8x - 3x - 2\sqrt{3}$$
Group the terms to factorize:
$$g(x) = 4x(\sqrt{3}x + 2) - \sqrt{3}(\sqrt{3}x + 2)$$
$$g(x) = (4x - \sqrt{3})(\sqrt{3}x + 2)$$
To find the zeroes, set $g(x) = 0$:
$$(4x - \sqrt{3})(\sqrt{3}x + 2) = 0$$
$$4x - \sqrt{3} = 0 \implies x = \frac{\sqrt{3}}{4}$$
$$\sqrt{3}x + 2 = 0 \implies x = -\frac{2}{\sqrt{3}}$$
Let $\alpha = \frac{\sqrt{3}}{4}$ and $\beta = -\frac{2}{\sqrt{3}}$.
Q3. If one zero of the polynomial $p(x) = (k^2 + 4)x^2 + 13x + 4k$ is the reciprocal of the other, find the value of $k$.
Answer: Value of $k$: $1$
Justification & Calculation:
Given polynomial: $p(x) = (k^2 + 4)x^2 + 13x + 4k$
Comparing this with the standard quadratic form $Ax^2 + Bx + C$, we get:
$$A = k^2 + 4, \quad B = 13, \quad C = 4k$$
Let one zero of the polynomial be $\alpha$. Since the other zero is the reciprocal of the first, the zeroes are $\alpha$ and $\frac{1}{\alpha}$.
Using Vieta's formulas, the product of the zeroes is given by $\frac{C}{A}$:
$$\alpha \cdot \frac{1}{\alpha} = \frac{4k}{k^2 + 4}$$
$$1 = \frac{4k}{k^2 + 4}$$
Cross-multiply to solve for $k$:
$$k^2 + 4 = 4k$$
$$k^2 - 4k + 4 = 0$$
This is a perfect square trinomial:
$$(k - 2)^2 = 0$$
$$k - 2 = 0$$
$$k = 2$$
Wait, let me double check the calculation:
If $k = 2$, $C = 4(2) = 8$, $A = 2^2 + 4 = 8$, product $\frac{8}{8} = 1$. Correct.
Wait, let me re-verify: $4k / (k^2+4) = 1 \implies k^2 - 4k + 4 = 0 \implies (k-2)^2 = 0 \implies k = 2$.
Let me correct the final value in the answer text to $2$ instead of $1$.
Value of $k$: $2$
Justification & Calculation:
Let the given polynomial be $p(x) = (k^2 + 4)x^2 + 13x + 4k$.
Comparing this with the standard quadratic form $Ax^2 + Bx + C$, we get:
$$A = k^2 + 4, \quad B = 13, \quad C = 4k$$
Let the zeroes be $\alpha$ and $\frac{1}{\alpha}$. Using the product of zeroes:
$$\alpha \cdot \frac{1}{\alpha} = \frac{4k}{k^2 + 4}$$
$$1 = \frac{4k}{k^2 + 4}$$
$$k^2 + 4 = 4k$$
$$k^2 - 4k + 4 = 0$$
$$(k - 2)^2 = 0$$
$$k = 2$$
Therefore, the value of $k$ is $2$.
Q4. If one zero of the quadratic polynomial $f(x) = 4x^2 - 8kx - 9$ is negative of the other, find the value of $k$ and write the resulting polynomial.
Answer: Value of $k$: $0$ Resulting polynomial: $4x^2 - 9$
Justification & Calculation:
Given polynomial: $f(x) = 4x^2 - 8kx - 9$
Comparing this with the standard quadratic form $ax^2 + bx + c$, we get:
$$a = 4, \quad b = -8k, \quad c = -9$$
Let one zero of the polynomial be $\alpha$. Since the other zero is the negative of the first, the zeroes are $\alpha$ and $-\alpha$.
Using Vieta's formulas, the sum of the zeroes is given by $-\frac{b}{a}$:
$$\alpha + (-\alpha) = -\frac{-8k}{4}$$
$$0 = \frac{8k}{4}$$
$$0 = 2k$$
$$k = 0$$
Now, substitute $k = 0$ back into the original polynomial to find the resulting polynomial:
$$f(x) = 4x^2 - 8(0)x - 9$$
$$f(x) = 4x^2 - 9$$
Therefore, the value of $k$ is $0$ and the resulting polynomial is $4x^2 - 9$.
Q5. Find the zeroes of the quadratic polynomial $h(x) = 2x^2 - (1 + 2\sqrt{2})x + \sqrt{2}$ and verify the relationship between coefficients and zeroes.
Answer: Zeroes of the polynomial: $\frac{1}{2}$ and $\sqrt{2}$
Justification & Calculation:
Given polynomial: $h(x) = 2x^2 - (1 + 2\sqrt{2})x + \sqrt{2}$
Comparing this with the standard quadratic form $ax^2 + bx + c$, we get:
$$a = 2, \quad b = -(1 + 2\sqrt{2}), \quad c = \sqrt{2}$$
Step 1: Find the zeroes by factorisation
We need to split the middle term $-(1 + 2\sqrt{2})x$ into two parts whose sum is $-(1 + 2\sqrt{2})$ and whose product is equal to $a \cdot c$:
$$a \cdot c = 2 \cdot \sqrt{2} = 2\sqrt{2}$$
The two numbers whose product is $2\sqrt{2}$ and whose sum is $-(1 + 2\sqrt{2})$ are $-1$ and $-2\sqrt{2}$.
Rewrite the middle term using $-1x$ and $-2\sqrt{2}x$:
$$h(x) = 2x^2 - 2\sqrt{2}x - 1x + \sqrt{2}$$
Group the terms to factorize:
$$h(x) = 2x(x - \sqrt{2}) - 1(x - \sqrt{2})$$
$$h(x) = (2x - 1)(x - \sqrt{2})$$
To find the zeroes, set $h(x) = 0$:
$$(2x - 1)(x - \sqrt{2}) = 0$$
$$2x - 1 = 0 \implies x = \frac{1}{2}$$
$$x - \sqrt{2} = 0 \implies x = \sqrt{2}$$
Let $\alpha = \frac{1}{2}$ and $\beta = \sqrt{2}$.
Q6. If the sum of the zeroes of the quadratic polynomial $f(t) = kt^2 + 2t + 3k$ is equal to their product, find the value of $k$.
Answer: Value of $k$: $-\frac{2}{3}$
Justification & Calculation:
Let the given quadratic polynomial be $f(t) = kt^2 + 2t + 3k$.
Comparing this with the standard quadratic form $at^2 + bt + c$, we get:
$$a = k, \quad b = 2, \quad c = 3k$$
Using Vieta's formulas for the zeroes of the polynomial:
$$\text{Sum of zeroes} = -\frac{b}{a} = -\frac{2}{k}$$
$$\text{Product of zeroes} = \frac{c}{a} = \frac{3k}{k} = 3$$
According to the given condition, the sum of the zeroes is equal to their product:
$$-\frac{2}{k} = 3$$
Cross-multiply to solve for $k$:
$$-2 = 3k$$
$$k = -\frac{2}{3}$$
Therefore, the value of $k$ is $-\frac{2}{3}$.
Q7. If $\alpha$ and $\beta$ are the zeroes of the polynomial $p(x) = 3x^2 - 6x + 4$, find the value of $k$ if it is given that $3\alpha + 2\beta = 4$.
Answer: Value of $k$ (or note on typo): Since there is no parameter $k$ in the given polynomial, the condition $3\alpha + 2\beta = 4$ is used to verify consistency, or find that the zeroes are $\alpha = 2 + \frac{2}{\sqrt{3}}i$ and $\beta = 2 - \frac{2}{\sqrt{3}}i$ (non-real). Wait, let's solve the system of equations.
Justification & Calculation:
Given polynomial: $p(x) = 3x^2 - 6x + 4$
Comparing with the standard quadratic form $ax^2 + bx + c$, we get:
$$a = 3, \quad b = -6, \quad c = 4$$
Using Vieta's formulas for the zeroes $\alpha$ and $\beta$:
$$\alpha + \beta = -\frac{b}{a} = -\frac{-6}{3} = 2 \quad \text{--- (Equation 1)}$$
$$\alpha\beta = \frac{c}{a} = \frac{4}{3} \quad \text{--- (Equation 2)}$$
We are also given the linear relation between the zeroes:
$$3\alpha + 2\beta = 4 \quad \text{--- (Equation 3)}$$
We can solve Equations (1) and (3) simultaneously to find the values of $\alpha$ and $\beta$:
From Equation (1), multiply by $2$:
$$2\alpha + 2\beta = 4 \quad \text{--- (Equation 4)}$$
Subtract Equation (4) from Equation (3):
$$(3\alpha + 2\beta) - (2\alpha + 2\beta) = 4 - 4$$
$$\alpha = 0$$
Substitute $\alpha = 0$ into Equation (1):
$$0 + \beta = 2 \implies \beta = 2$$
Now, check these values using Equation (2) (Product of zeroes):
$$\alpha\beta = (0)(2) = 0$$
However, from Vieta's formula, the product of zeroes is $\frac{4}{3}$, which means $0 \neq \frac{4}{3}$.
This contradiction implies that there might be a typo in the question statement or coefficient (often such questions involve a parameter $k$, e.g., $3x^2 - kx + 4$, or the linear relation is different). If the polynomial is fixed as $3x^2 - 6x + 4$, the zeroes are complex numbers and cannot satisfy the given real linear equation while matching Vieta's product.
*(Note: If the question meant to find a parameter $k$ in the polynomial, such as $3x^2 - kx + 4$, let's re-evaluate assuming $b = -k$):*
$$\alpha + \beta = \frac{k}{3}$$
Using $\alpha = 0, \beta = 2$:
$$\alpha + \beta = 2 \implies \frac{k}{3} = 2 \implies k = 6$$
And product $\alpha\beta = 0 \neq \frac{4}{3}$, so the polynomial coefficients as given in the standard textbook problem usually feature a variable like $kx^2$ or $-kx$. Assuming the question as written contains a classic inconsistency or typographical error from standard question banks where $6$ is replaced by $k$ (i.e., $3x^2 - kx + 4$), the value of $k$ would be $6$.
Q8. Find the values of $a$ and $b$ if the zeroes of the polynomial $x^2 + ax - b$ are $3$ and $-4$.
Answer: Values: $a = 1$, $b = 12$
Justification & Calculation:
Let the given polynomial be $p(x) = x^2 + ax - b$.
Comparing this with the standard quadratic form $Ax^2 + Bx + C$, we get:
$$A = 1, \quad B = a, \quad C = -b$$
We are given that the zeroes of the polynomial are $3$ and $-4$.
Let $\alpha = 3$ and $\beta = -4$.
Using Vieta's formulas:
1. **Sum of zeroes:**
$$\alpha + \beta = -\frac{B}{A}$$
$$3 + (-4) = -\frac{a}{1}$$
$$-1 = -a$$
$$a = 1$$
2. **Product of zeroes:**
$$\alpha\beta = \frac{C}{A}$$
$$(3)(-4) = \frac{-b}{1}$$
$$-12 = -b$$
$$b = 12$$
Therefore, the values of $a$ and $b$ are $1$ and $12$ respectively.
Q9. If $\alpha$ and $\beta$ are the zeroes of the quadratic polynomial $f(x) = x^2 - kx + 6$ such that $(\alpha + \beta)^2 - 2\alpha\beta = 13$, find the value of $k$.
Answer: Value of $k$: $k = \pm 5$
Justification & Calculation:
Given quadratic polynomial: $f(x) = x^2 - kx + 6$
Comparing this with the standard quadratic form $ax^2 + bx + c$, we get:
$$a = 1, \quad b = -k, \quad c = 6$$
Using Vieta's formulas for the zeroes $\alpha$ and $\beta$:
$$\alpha + \beta = -\frac{b}{a} = -\frac{-k}{1} = k$$
$$\alpha\beta = \frac{c}{a} = \frac{6}{1} = 6$$
We are given the condition:
$$(\alpha + \beta)^2 - 2\alpha\beta = 13$$
Substitute the values from Vieta's formulas into the condition:
$$(k)^2 - 2(6) = 13$$
$$k^2 - 12 = 13$$
$$k^2 = 13 + 12$$
$$k^2 = 25$$
$$k = \pm\sqrt{25}$$
$$k = \pm 5$$
Therefore, the value of $k$ is $5$ or $-5$.
Q10. If the product of the zeroes of the polynomial $g(x) = ax^2 - 6x - 6$ is $4$, find the value of $a$. Hence, find the sum of its zeroes.
Answer: Value of $a$: $-\frac{3}{2}$ Sum of zeroes: $-4$
Justification & Calculation:
Given polynomial: $g(x) = ax^2 - 6x - 6$
Comparing this with the standard quadratic form $Ax^2 + Bx + C$, we get:
$$A = a, \quad B = -6, \quad C = -6$$
Step 1: Find the value of $a$
Using Vieta's formula, the product of the zeroes is given by $\frac{C}{A}$:
$$\text{Product of zeroes} = \frac{-6}{a}$$
We are given that the product of the zeroes is $4$:
$$\frac{-6}{a} = 4$$
Solve for $a$:
$$4a = -6$$
$$a = \frac{-6}{4} = -\frac{3}{2}$$
Step 2: Find the sum of the zeroes
Using Vieta's formula, the sum of the zeroes is given by $-\frac{B}{A}$:
$$\text{Sum of zeroes} = -\frac{-6}{a} = \frac{6}{a}$$
Substitute the value of $a = -\frac{3}{2}$:
$$\text{Sum of zeroes} = \frac{6}{-\frac{3}{2}} = 6 \cdot \left(-\frac{2}{3}\right) = -4$$
Therefore, the value of $a$ is $-\frac{3}{2}$ and the sum of its zeroes is $-4$.
Q11. If the zeroes of the polynomial $f(x) = 2x^2 - 7x + 3$ are $\alpha$ and $\beta$, verify whether $\alpha + \beta = \frac{7}{2}$ and $\alpha\beta = \frac{3}{2}$ by computing the zeroes directly.
Answer: Zeroes computed directly: $\frac{1}{2}$ and $3$ Verification: $\alpha + \beta = \frac{7}{2}$ and $\alpha\beta = \frac{3}{2}$ are verified.
Justification & Calculation:
Given polynomial: $f(x) = 2x^2 - 7x + 3$
Step 1: Compute the zeroes directly by factorisation
We need to split the middle term $-7x$ into two parts whose sum is $-7$ and whose product is $2 \times 3 = 6$. These numbers are $-6$ and $-1$.
$$f(x) = 2x^2 - 6x - 1x + 3$$
$$f(x) = 2x(x - 3) - 1(x - 3)$$
$$f(x) = (2x - 1)(x - 3)$$
To find the zeroes, set $f(x) = 0$:
$$(2x - 1)(x - 3) = 0$$
$$2x - 1 = 0 \implies x = \frac{1}{2}$$
$$x - 3 = 0 \implies x = 3$$
Let $\alpha = \frac{1}{2}$ and $\beta = 3$.
Step 2: Verify the given relations
* **Sum of the zeroes ($\alpha + \beta$):**
$$\alpha + \beta = \frac{1}{2} + 3 = \frac{1 + 6}{2} = \frac{7}{2}$$
This matches the given value $\frac{7}{2}$. $\quad (\text{Verified})$
* **Product of the zeroes ($\alpha\beta$):**
$$\alpha\beta = \left(\frac{1}{2}\right)(3) = \frac{3}{2}$$
This matches the given value $\frac{3}{2}$. $\quad (\text{Verified})$
Q12. If one zero of the polynomial $p(x) = 3x^2 - kx - 5$ is $-1$, find the value of $k$. Using this value, find the other zero.
Answer: Value of $k$: $-2$ Other zero: $\frac{5}{3}$
Justification & Calculation:
Given polynomial: $p(x) = 3x^2 - kx - 5$
Step 1: Find the value of $k$
Since $-1$ is a zero of the polynomial $p(x)$, substituting $x = -1$ must make the polynomial equal to zero:
$$p(-1) = 3(-1)^2 - k(-1) - 5 = 0$$
$$3(1) + k - 5 = 0$$
$$3 + k - 5 = 0$$
$$k - 2 = 0$$
$$k = 2$$
Wait, let me double-check the sign:
$3(-1)^2 - k(-1) - 5 = 3(1) + k - 5 = k - 2 = 0 \implies k = 2$.
Let me correct the value of $k$ to $2$ in the answer block.
Value of $k$: $2$ Other zero: $\frac{5}{3}$
Justification & Calculation:
Since $-1$ is a zero of the polynomial $p(x) = 3x^2 - kx - 5$, we have:
$$p(-1) = 3(-1)^2 - k(-1) - 5 = 0$$
$$3 + k - 5 = 0$$
$$k - 2 = 0 \implies k = 2$$
Step 2: Find the other zero
Substitute $k = 2$ back into the polynomial:
$$p(x) = 3x^2 - 2x - 5$$
Let the zeroes be $\alpha = -1$ and $\beta$ (the other zero). Using Vieta's formula for the product of zeroes ($\alpha\beta = \frac{c}{a}$):
$$(-1) \cdot \beta = \frac{-5}{3}$$
$$-\beta = -\frac{5}{3}$$
$$\beta = \frac{5}{3}$$
Alternatively, using the sum of zeroes ($\alpha + \beta = -\frac{b}{a}$):
$$-1 + \beta = -\frac{-2}{3} = \frac{2}{3}$$
$$\beta = \frac{2}{3} + 1 = \frac{5}{3}$$
Therefore, the value of $k$ is $2$ and the other zero is $\frac{5}{3}$.
Q13. Find the value of $p$ for which the quadratic polynomial $4x^2 - px + 1$ has two equal zeroes.
Answer: Value of $p$: $p = 4$ or $p = -4$
Justification & Calculation:
Given quadratic polynomial: $f(x) = 4x^2 - px + 1$
Comparing this with the standard quadratic form $ax^2 + bx + c$, we get:
$$a = 4, \quad b = -p, \quad c = 1$$
For a quadratic polynomial to have two equal zeroes, its discriminant ($D$) must be equal to zero:
$$D = b^2 - 4ac = 0$$
Substitute the values of $a$, $b$, and $c$ into the discriminant formula:
$$(-p)^2 - 4(4)(1) = 0$$
$$p^2 - 16 = 0$$
$$p^2 = 16$$
$$p = \pm\sqrt{16}$$
$$p = \pm 4$$
Therefore, the value of $p$ is $4$ or $-4$.
Q14. If the sum of the zeroes of a quadratic polynomial $ax^2 + bx + c$ is $0$ and one zero is $3$, find the values of $b$ and $c$ in terms of $a$.
Answer: Values: $b = 0$, $c = -9a$
Justification & Calculation:
Given quadratic polynomial: $p(x) = ax^2 + bx + c$
Let the zeroes of the polynomial be $\alpha$ and $\beta$.
We are given:
* Sum of the zeroes: $\alpha + \beta = 0$
* One zero: $\alpha = 3$
Step 1: Find the other zero ($\beta$)
Substitute $\alpha = 3$ into the sum of zeroes equation:
$$3 + \beta = 0$$
$$\beta = -3$$
Step 2: Find $b$ in terms of $a$
Using Vieta's formula for the sum of zeroes:
$$\text{Sum of zeroes} = -\frac{b}{a}$$
$$3 + (-3) = -\frac{b}{a}$$
$$0 = -\frac{b}{a}$$
$$b = 0$$
Step 3: Find $c$ in terms of $a$
Using Vieta's formula for the product of zeroes:
$$\text{Product of zeroes} = \frac{c}{a}$$
$$\alpha\beta = \frac{c}{a}$$
$$(3)(-3) = \frac{c}{a}$$
$$-9 = \frac{c}{a}$$
$$c = -9a$$
Therefore, the values in terms of $a$ are $b = 0$ and $c = -9a$.
Q15. If $\alpha$ and $\beta$ are the zeroes of the polynomial $x^2 - 5x + k$ such that $\alpha - \beta = 1$, find the value of $k$.
Answer: Value of $k$: $6$
Justification & Calculation:
Given polynomial: $p(x) = x^2 - 5x + k$
Comparing this with the standard quadratic form $ax^2 + bx + c$, we get:
$$a = 1, \quad b = -5, \quad c = k$$
Using Vieta's formulas for the zeroes $\alpha$ and $\beta$:
$$\alpha + \beta = -\frac{b}{a} = -\frac{-5}{1} = 5 \quad \text{--- (Equation 1)}$$
$$\alpha\beta = \frac{c}{a} = \frac{k}{1} = k \quad \text{--- (Equation 2)}$$
We are given the relation between the zeroes:
$$\alpha - \beta = 1 \quad \text{--- (Equation 3)}$$
Step 1: Find the values of $\alpha$ and $\beta$
Add Equation (1) and Equation (3) to solve for $\alpha$:
$$(\alpha + \beta) + (\alpha - \beta) = 5 + 1$$
$$2\alpha = 6$$
$$\alpha = 3$$
Substitute $\alpha = 3$ into Equation (1) to solve for $\beta$:
$$3 + \beta = 5$$
$$\beta = 2$$
Step 2: Find the value of $k$
Substitute $\alpha = 3$ and $\beta = 2$ into Equation (2) (Product of zeroes):
$$\alpha\beta = k$$
$$(3)(2) = k$$
$$k = 6$$
Therefore, the value of $k$ is $6$.
Q16. If $\alpha$ and $\beta$ are the zeroes of the quadratic polynomial $f(x) = x^2 - 4x + 3$, find the value of $\alpha^4\beta^3 + \alpha^3\beta^4$.
Answer: Value: $108$
Justification & Calculation:
Given quadratic polynomial: $f(x) = x^2 - 4x + 3$
Comparing this with the standard quadratic form $ax^2 + bx + c$, we get:
$$a = 1, \quad b = -4, \quad c = 3$$
Using Vieta's formulas for the zeroes $\alpha$ and $\beta$:
$$\alpha + \beta = -\frac{b}{a} = -\frac{-4}{1} = 4$$
$$\alpha\beta = \frac{c}{a} = \frac{3}{1} = 3$$
Step 1: Simplify the expression
We need to find the value of $\alpha^4\beta^3 + \alpha^3\beta^4$.
Factor out the common term $\alpha^3\beta^3$:
$$\alpha^4\beta^3 + \alpha^3\beta^4 = \alpha^3\beta^3(\alpha + \beta)$$
This can also be written as:
$$(\alpha\beta)^3(\alpha + \beta)$$
Step 2: Substitute the values from Vieta's formulas
$$(3)^3 \cdot (4) = 27 \cdot 4 = 108$$
Therefore, the value of $\alpha^4\beta^3 + \alpha^3\beta^4$ is $108$.
Q17. If $\alpha$ and $\beta$ are the zeroes of the polynomial $p(x) = 2x^2 + 5x + k$ such that $\alpha^2 + \beta^2 + \alpha\beta = \frac{21}{4}$, find the value of $k$.
Answer: Value of $k$: $2$
Justification & Calculation:
Given polynomial: $p(x) = 2x^2 + 5x + k$
Comparing this with the standard quadratic form $ax^2 + bx + c$, we get:
$$a = 2, \quad b = 5, \quad c = k$$
Using Vieta's formulas for the zeroes $\alpha$ and $\beta$:
$$\alpha + \beta = -\frac{b}{a} = -\frac{5}{2}$$
$$\alpha\beta = \frac{c}{a} = \frac{k}{2}$$
Step 1: Relate the given expression to sum and product of zeroes
We are given the condition:
$$\alpha^2 + \beta^2 + \alpha\beta = \frac{21}{4}$$
Rewrite $\alpha^2 + \beta^2$ using the identity $(\alpha + \beta)^2 = \alpha^2 + \beta^2 + 2\alpha\beta$, which gives $\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta$:
$$[(\alpha + \beta)^2 - 2\alpha\beta] + \alpha\beta = \frac{21}{4}$$
$$(\alpha + \beta)^2 - \alpha\beta = \frac{21}{4}$$
Step 2: Substitute the values from Vieta's formulas
$$\left(-\frac{5}{2}\right)^2 - \frac{k}{2} = \frac{21}{4}$$
$$\frac{25}{4} - \frac{k}{2} = \frac{21}{4}$$
Rearrange the equation to solve for $k$:
$$\frac{25}{4} - \frac{21}{4} = \frac{k}{2}$$
$$\frac{4}{4} = \frac{k}{2}$$
$$1 = \frac{k}{2}$$
$$k = 2$$
Therefore, the value of $k$ is $2$.
Q18. If $\alpha$ and $\beta$ are the zeroes of the quadratic polynomial $g(x) = x^2 - px + q$, find the value of:
(a) $\frac{1}{\alpha} + \frac{1}{\beta}$
(b) $\alpha^2 + \beta^2$
Answer: (a) $\frac{p}{q}$ (b) $p^2 - 2q$
Justification & Calculation:
Given quadratic polynomial: $g(x) = x^2 - px + q$
Comparing this with the standard quadratic form $ax^2 + bx + c$, we get:
$$a = 1, \quad b = -p, \quad c = q$$
Using Vieta's formulas for the zeroes $\alpha$ and $\beta$:
$$\alpha + \beta = -\frac{b}{a} = -\frac{-p}{1} = p$$
$$\alpha\beta = \frac{c}{a} = \frac{q}{1} = q$$
(a) Find the value of $\frac{1}{\alpha} + \frac{1}{\beta}$
Take the common denominator to simplify the expression:
$$\frac{1}{\alpha} + \frac{1}{\beta} = \frac{\beta + \alpha}{\alpha\beta} = \frac{\alpha + \beta}{\alpha\beta}$$
Substitute the values from Vieta's formulas:
$$\frac{1}{\alpha} + \frac{1}{\beta} = \frac{p}{q}$$
(b) Find the value of $\alpha^2 + \beta^2$
Use the algebraic identity $(\alpha + \beta)^2 = \alpha^2 + \beta^2 + 2\alpha\beta$, which can be rearranged as:
$$\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta$$
Substitute the values from Vieta's formulas:
$$\alpha^2 + \beta^2 = (p)^2 - 2(q) = p^2 - 2q$$
Therefore, the values are $\frac{p}{q}$ for part (a) and $p^2 - 2q$ for part (b).
Q19. If $\alpha$ and $\beta$ are the zeroes of the quadratic polynomial $f(x) = 3x^2 - 5x - 2$, evaluate the value of $\frac{\alpha^2}{\beta} + \frac{\beta^2}{\alpha}$.
Answer: Value: $-\frac{215}{18}$
Justification & Calculation:
Given quadratic polynomial: $f(x) = 3x^2 - 5x - 2$
Comparing this with the standard quadratic form $ax^2 + bx + c$, we get:
$$a = 3, \quad b = -5, \quad c = -2$$
Using Vieta's formulas for the zeroes $\alpha$ and $\beta$:
$$\alpha + \beta = -\frac{b}{a} = -\frac{-5}{3} = \frac{5}{3}$$
$$\alpha\beta = \frac{c}{a} = \frac{-2}{3}$$
Step 1: Simplify the expression
We need to evaluate $\frac{\alpha^2}{\beta} + \frac{\beta^2}{\alpha}$.
Find a common denominator:
$$\frac{\alpha^2}{\beta} + \frac{\beta^2}{\alpha} = \frac{\alpha^3 + \beta^3}{\alpha\beta}$$
Use the algebraic identity for the sum of cubes: $\alpha^3 + \beta^3 = (\alpha + \beta)(\alpha^2 - \alpha\beta + \beta^2)$ or $(\alpha + \beta)^3 - 3\alpha\beta(\alpha + \beta)$. Let's use the latter form:
$$\frac{(\alpha + \beta)^3 - 3\alpha\beta(\alpha + \beta)}{\alpha\beta}$$
Step 2: Substitute the values from Vieta's formulas
Numerator:
$$(\alpha + \beta)^3 - 3\alpha\beta(\alpha + \beta) = \left(\frac{5}{3}\right)^3 - 3\left(-\frac{2}{3}\right)\left(\frac{5}{3}\right)$$
$$= \frac{125}{27} + \frac{30}{9} = \frac{125}{27} + \frac{90}{27} = \frac{215}{27}$$
Denominator:
$$\alpha\beta = -\frac{2}{3}$$
Now, put them together:
$$\frac{\frac{215}{27}}{-\frac{2}{3}} = \frac{215}{27} \cdot \left(-\frac{3}{2}\right) = \frac{215}{9} \cdot \left(-\frac{1}{2}\right) = -\frac{215}{18}$$
Therefore, the value of $\frac{\alpha^2}{\beta} + \frac{\beta^2}{\alpha}$ is $-\frac{215}{18}$.
Q20. If $\alpha$ and $\beta$ are the zeroes of the polynomial $p(x) = x^2 - x - 2$, find the value of $\frac{1}{2\alpha + 1} + \frac{1}{2\beta + 1}$.
Answer: Value: $\frac{3}{5}$
Justification & Calculation:
Given polynomial: $p(x) = x^2 - x - 2$
Comparing this with the standard quadratic form $ax^2 + bx + c$, we get:
$$a = 1, \quad b = -1, \quad c = -2$$
Using Vieta's formulas for the zeroes $\alpha$ and $\beta$:
$$\alpha + \beta = -\frac{b}{a} = -\frac{-1}{1} = 1$$
$$\alpha\beta = \frac{c}{a} = \frac{-2}{1} = -2$$
Step 1: Simplify the expression
We need to find the value of:
$$\frac{1}{2\alpha + 1} + \frac{1}{2\beta + 1}$$
Find a common denominator:
$$\frac{(2\beta + 1) + (2\alpha + 1)}{(2\alpha + 1)(2\beta + 1)} = \frac{2(\alpha + \beta) + 2}{4\alpha\beta + 2(\alpha + \beta) + 1}$$
Step 2: Substitute the values from Vieta's formulas
* **Numerator:**
$$2(\alpha + \beta) + 2 = 2(1) + 2 = 2 + 2 = 4$$
* **Denominator:**
$$4\alpha\beta + 2(\alpha + \beta) + 1 = 4(-2) + 2(1) + 1 = -8 + 2 + 1 = -5$$
Now, put them together:
$$\frac{4}{-5} = -\frac{4}{5}$$
Wait, let me re-verify the direct computation of zeroes to double-check:
$p(x) = x^2 - x - 2 = (x - 2)(x + 1) = 0 \implies \alpha = 2, \beta = -1$ (or vice versa).
Let's calculate using the direct zeroes:
$$\frac{1}{2(2) + 1} + \frac{1}{2(-1) + 1} = \frac{1}{4 + 1} + \frac{1}{-2 + 1} = \frac{1}{5} + \frac{1}{-1} = \frac{1}{5} - 1 = -\frac{4}{5}$$
Ah, let's fix the answer value from $\frac{3}{5}$ to $-\frac{4}{5}$.
Answer: Value: $-\frac{4}{5}$
Justification & Calculation:
Given polynomial: $p(x) = x^2 - x - 2$
Comparing this with the standard quadratic form $ax^2 + bx + c$, we get:
$$a = 1, \quad b = -1, \quad c = -2$$
Using Vieta's formulas for the zeroes $\alpha$ and $\beta$:
$$\alpha + \beta = -\frac{b}{a} = -\frac{-1}{1} = 1$$
$$\alpha\beta = \frac{c}{a} = \frac{-2}{1} = -2$$
Step 1: Simplify the expression
$$\frac{1}{2\alpha + 1} + \frac{1}{2\beta + 1} = \frac{(2\beta + 1) + (2\alpha + 1)}{(2\alpha + 1)(2\beta + 1)}$$
$$= \frac{2(\alpha + \beta) + 2}{4\alpha\beta + 2(\alpha + \beta) + 1}$$
Step 2: Substitute the values from Vieta's formulas
Numerator:
$$2(1) + 2 = 4$$
Denominator:
$$4(-2) + 2(1) + 1 = -8 + 2 + 1 = -5$$
Result:
$$\frac{4}{-5} = -\frac{4}{5}$$
Therefore, the value of the expression is $-\frac{4}{5}$.
Q21. If $\alpha$ and $\beta$ are the zeroes of the polynomial $f(x) = 2x^2 - 5x + 7$, find a quadratic polynomial whose zeroes are $2\alpha + 3\beta$ and $3\alpha + 2\beta$.
Answer: Quadratic Polynomial: $x^2 - 10x + 28$ (or any polynomial of the form $k(x^2 - 10x + 28)$ where $k \neq 0$ is a real number)
Justification & Calculation:
Given polynomial: $f(x) = 2x^2 - 5x + 7$
Comparing this with the standard quadratic form $ax^2 + bx + c$, we get:
$$a = 2, \quad b = -5, \quad c = 7$$
Using Vieta's formulas for the zeroes $\alpha$ and $\beta$:
$$\alpha + \beta = -\frac{b}{a} = -\frac{-5}{2} = \frac{5}{2}$$
$$\alpha\beta = \frac{c}{a} = \frac{7}{2}$$
Step 1: Find the new zeroes
Let the new zeroes be $S_1$ and $S_2$:
$$S_1 = 2\alpha + 3\beta$$
$$S_2 = 3\alpha + 2\beta$$
Step 2: Find the sum of the new zeroes ($\text{Sum} = S_1 + S_2$)
$$\text{Sum} = (2\alpha + 3\beta) + (3\alpha + 2\beta) = 5\alpha + 5\beta = 5(\alpha + \beta)$$
Substitute the value of $\alpha + \beta$:
$$\text{Sum} = 5\left(\frac{5}{2}\right) = \frac{25}{2}$$
Step 3: Find the product of the new zeroes ($\text{Product} = S_1 \cdot S_2$)
$$\text{Product} = (2\alpha + 3\beta)(3\alpha + 2\beta)$$
$$= 6\alpha^2 + 4\alpha\beta + 9\alpha\beta + 6\beta^2$$
$$= 6(\alpha^2 + \beta^2) + 13\alpha\beta$$
Recall the identity for $\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta$:
$$\text{Product} = 6[(\alpha + \beta)^2 - 2\alpha\beta] + 13\alpha\beta$$
$$= 6(\alpha + \beta)^2 - 12\alpha\beta + 13\alpha\beta$$
$$= 6(\alpha + \beta)^2 + \alpha\beta$$
Substitute the values of $\alpha + \beta = \frac{5}{2}$ and $\alpha\beta = \frac{7}{2}$:
$$\text{Product} = 6\left(\frac{5}{2}\right)^2 + \frac{7}{2}$$
$$= 6\left(\frac{25}{4}\right) + \frac{7}{2}$$
$$= \frac{150}{4} + \frac{7}{2} = \frac{75}{2} + \frac{7}{2} = \frac{82}{2} = 41$$
Wait, let's re-verify the sum of the new zeroes:
$\text{Sum} = 5(\alpha + \beta) = 5\left(\frac{5}{2}\right) = \frac{25}{2}$.
Wait, standard problems like this usually have nice integer coefficients. Let's check if the polynomial values or question text from standard banks has a different constant, or if fractions are correct.
Let's check: if polynomial is $x^2 - (\text{Sum})x + \text{Product} = x^2 - \frac{25}{2}x + 41$, multiplying by $2$ gives $2\left(x^2 - \frac{25}{2}x + 41\right) = 2x^2 - 25x + 82$.
Let's present the polynomial using the standard form $k(x^2 - \text{Sum}\cdot x + \text{Product})$ or with fractions. Let's write it cleanly with fractions or scaled by $2$.
Let's write:
A quadratic polynomial is given by $P(x) = x^2 - (\text{Sum})x + \text{Product}$:
$$P(x) = x^2 - \frac{25}{2}x + 41$$
Multiplying by $2$ to clear fractions (or expressing as $k(2x^2 - 25x + 82)$):
$$P(x) = 2x^2 - 25x + 82$$
Therefore, the required quadratic polynomial is $2x^2 - 25x + 82$ (or $x^2 - \frac{25}{2}x + 41$).
Q22. If $\alpha$ and $\beta$ are the zeroes of the polynomial $x^2 - 6x + k$, find the value of $k$ if $3\alpha + 2\beta = 20$.
Answer: Value of $k$: $-16$
Justification & Calculation:
Given polynomial: $p(x) = x^2 - 6x + k$
Comparing this with the standard quadratic form $ax^2 + bx + c$, we get:
$$a = 1, \quad b = -6, \quad c = k$$
Using Vieta's formulas for the zeroes $\alpha$ and $\beta$:
$$\alpha + \beta = -\frac{b}{a} = -\frac{-6}{1} = 6 \quad \text{--- (Equation 1)}$$
$$\alpha\beta = \frac{c}{a} = \frac{k}{1} = k \quad \text{--- (Equation 2)}$$
We are also given the linear relation between the zeroes:
$$3\alpha + 2\beta = 20 \quad \text{--- (Equation 3)}$$
Step 1: Find the values of $\alpha$ and $\beta$
Multiply Equation (1) by $2$:
$$2\alpha + 2\beta = 12 \quad \text{--- (Equation 4)}$$
Subtract Equation (4) from Equation (3):
$$(3\alpha + 2\beta) - (2\alpha + 2\beta) = 20 - 12$$
$$\alpha = 8$$
Substitute $\alpha = 8$ into Equation (1):
$$8 + \beta = 6$$
$$\beta = 6 - 8 = -2$$
Step 2: Find the value of $k$
Substitute $\alpha = 8$ and $\beta = -2$ into Equation (2) (Product of zeroes):
$$\alpha\beta = k$$
$$(8)(-2) = k$$
$$k = -16$$
Therefore, the value of $k$ is $-16$.
Q23. If $\alpha, \beta$ are the zeroes of the quadratic polynomial $f(x) = x^2 - p(x + 1) - c$, show that $(\alpha + 1)(\beta + 1) = 1 - c$.
Justification & Calculation:
Given quadratic polynomial: $f(x) = x^2 - p(x + 1) - c$
Step 1: Simplify the polynomial into standard quadratic form
Expand the expression:
$$f(x) = x^2 - px - p - c$$
$$f(x) = x^2 - px + (-p - c)$$
Comparing this with the standard quadratic form $ax^2 + bx + d$, we get:
$$a = 1, \quad b = -p, \quad d = -p - c$$
Step 2: Use Vieta's formulas
Since $\alpha$ and $\beta$ are the zeroes of $f(x)$:
$$\alpha + \beta = -\frac{b}{a} = -\frac{-p}{1} = p$$
$$\alpha\beta = \frac{d}{a} = \frac{-p - c}{1} = -p - c$$
Step 3: Expand the given expression to prove the identity
Consider the left-hand side (LHS):
$$(\alpha + 1)(\beta + 1)$$
Multiply the terms:
$$= \alpha\beta + \alpha + \beta + 1$$
Substitute the values from Vieta's formulas ($\alpha + \beta = p$ and $\alpha\beta = -p - c$):
$$= (-p - c) + p + 1$$
$$= -p - c + p + 1$$
$$= 1 - c$$
Since LHS = RHS ($1 - c$), the identity is verified.
Q24. If $\alpha$ and $\beta$ are the zeroes of the polynomial $g(x) = 4x^2 - 5x - 1$, find the value of $\alpha^2\beta + \alpha\beta^2$.
Answer: Value: $-\frac{5}{16}$
Justification & Calculation:
Given quadratic polynomial: $g(x) = 4x^2 - 5x - 1$
Comparing this with the standard quadratic form $ax^2 + bx + c$, we get:
$$a = 4, \quad b = -5, \quad c = -1$$
Using Vieta's formulas for the zeroes $\alpha$ and $\beta$:
$$\alpha + \beta = -\frac{b}{a} = -\frac{-5}{4} = \frac{5}{4}$$
$$\alpha\beta = \frac{c}{a} = \frac{-1}{4} = -\frac{1}{4}$$
Step 1: Factorize the expression
We need to find the value of $\alpha^2\beta + \alpha\beta^2$.
Factor out the common terms $\alpha\beta$:
$$\alpha^2\beta + \alpha\beta^2 = \alpha\beta(\alpha + \beta)$$
Step 2: Substitute the values from Vieta's formulas
$$\alpha\beta(\alpha + \beta) = \left(-\frac{1}{4}\right) \cdot \left(\frac{5}{4}\right)$$
$$= -\frac{5}{16}$$
Therefore, the value of $\alpha^2\beta + \alpha\beta^2$ is $-\frac{5}{16}$.
Q25. If $\alpha$ and $\beta$ are the zeroes of the quadratic polynomial $p(x) = 2x^2 + 3x - 1$, find the value of $\frac{\alpha}{\beta} + \frac{\beta}{\alpha}$.
Answer: Value: $-\frac{13}{2}$
Justification & Calculation:
Given quadratic polynomial: $p(x) = 2x^2 + 3x - 1$
Comparing this with the standard quadratic form $ax^2 + bx + c$, we get:
$$a = 2, \quad b = 3, \quad c = -1$$
Using Vieta's formulas for the zeroes $\alpha$ and $\beta$:
$$\alpha + \beta = -\frac{b}{a} = -\frac{3}{2}$$
$$\alpha\beta = \frac{c}{a} = \frac{-1}{2}$$
Step 1: Simplify the expression
We need to find the value of $\frac{\alpha}{\beta} + \frac{\beta}{\alpha}$.
Find a common denominator:
$$\frac{\alpha}{\beta} + \frac{\beta}{\alpha} = \frac{\alpha^2 + \beta^2}{\alpha\beta}$$
Recall the algebraic identity for $\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta$:
$$\frac{(\alpha + \beta)^2 - 2\alpha\beta}{\alpha\beta}$$
Step 2: Substitute the values from Vieta's formulas
Numerator:
$$(\alpha + \beta)^2 - 2\alpha\beta = \left(-\frac{3}{2}\right)^2 - 2\left(-\frac{1}{2}\right)$$
$$= \frac{9}{4} + 1 = \frac{9}{4} + \frac{4}{4} = \frac{13}{4}$$
Denominator:
$$\alpha\beta = -\frac{1}{2}$$
Now, put them together:
$$\frac{\frac{13}{4}}{-\frac{1}{2}} = \frac{13}{4} \cdot \left(-\frac{2}{1}\right) = -\frac{13}{2}$$
Therefore, the value of $\frac{\alpha}{\beta} + \frac{\beta}{\alpha}$ is $-\frac{13}{2}$.
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.
