CBSE Class 10 Maths Chapter 2 Polynomials Model Questions - 2 Marks - Part 2
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CBSE Class 10 Maths Chapter 2 Polynomials Model Questions - 2 Marks - Part 2
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SECTION B — Very Short Answer Type Questions
[2 Marks Each]
Q26. Find the zeroes of the quadratic polynomial $p(x) = x^2 + 2x - 15$.
Answer: Zeroes of the polynomial: $3$ and $-5$
Justification & Calculation:
To find the zeroes, set $p(x) = 0$:
$$x^2 + 2x - 15 = 0$$
Using the splitting the middle term method (find two numbers that multiply to $-15$ and add up to $2$, which are $5$ and $-3$):
$$x^2 + 5x - 3x - 15 = 0$$
$$x(x + 5) - 3(x + 5) = 0$$
$$(x - 3)(x + 5) = 0$$
Setting each factor to zero:
$$x - 3 = 0 \implies x = 3$$
$$x + 5 = 0 \implies x = -5$$
Therefore, the zeroes of the polynomial are $3$ and $-5$.
Q27. Find the value of $x$ for which the polynomial $p(x) = 3x^2 - 12$ vanishes.
Answer: Values of $x$: $2$ and $-2$
Justification & Calculation:
A polynomial "vanishes" where its value is equal to zero. To find these values, set $p(x) = 0$:
$$3x^2 - 12 = 0$$
Divide the entire equation by $3$:
$$x^2 - 4 = 0$$
$$x^2 = 4$$
Taking the square root on both sides:
$$x = \pm 2$$
Therefore, the polynomial vanishes at $x = 2$ and $x = -2$.
Q28. Find the zeroes of the polynomial $f(x) = 5x^2 - 20x$.
Answer: Zeroes of the polynomial: $0$ and $4$
Justification & Calculation:
To find the zeroes, set $f(x) = 0$:
$$5x^2 - 20x = 0$$
Factor out the common term $5x$:
$$5x(x - 4) = 0$$
Setting each factor to zero:
$$5x = 0 \implies x = 0$$
$$x - 4 = 0 \implies x = 4$$
Therefore, the zeroes of the polynomial are $0$ and $4$.
Q29. If $\alpha$ and $\beta$ are the zeroes of the quadratic polynomial $p(x) = x^2 - 5x + 6$, find the value of $\alpha + \beta$ and $\alpha\beta$.
Answer: Sum of zeroes ($\alpha + \beta$): $5$ Product of zeroes ($\alpha\beta$): $6$
Justification & Calculation:
For a general quadratic polynomial $ax^2 + bx + c$, the relationships between zeroes and coefficients (Vieta's formulas) are given by:
$$\alpha + \beta = -\frac{b}{a}$$
$$\alpha\beta = \frac{c}{a}$$
Comparing the given polynomial $p(x) = x^2 - 5x + 6$ with $ax^2 + bx + c$, we get:
$$a = 1, \quad b = -5, \quad c = 6$$
Substituting these values into the formulas:
$$\alpha + \beta = -\frac{-5}{1} = 5$$
$$\alpha\beta = \frac{6}{1} = 6$$
Therefore, $\alpha + \beta = 5$ and $\alpha\beta = 6$.
Q30. For the polynomial $f(x) = 3x^2 + 5x - 2$, verify the relation between the sum of zeroes and its coefficients.
Answer: Sum of zeroes from roots: $-\frac{5}{3}$ Sum of zeroes from coefficients ($\frac{-b}{a}$): $-\frac{5}{3}$ (Verified)
Justification & Calculation:
First, find the zeroes of $f(x) = 3x^2 + 5x - 2$ by setting $f(x) = 0$:
$$3x^2 + 5x - 2 = 0$$
Using the splitting the middle term method (two numbers multiplying to $3 \times (-2) = -6$ and adding to $5$, which are $6$ and $-1$):
$$3x^2 + 6x - x - 2 = 0$$
$$3x(x + 2) - 1(x + 2) = 0$$
$$(3x - 1)(x + 2) = 0$$
Setting each factor to zero gives the zeroes:
$$x = \frac{1}{3} \quad \text{and} \quad x = -2$$
1. **Sum of zeroes:**
$$\text{Sum} = \frac{1}{3} + (-2) = \frac{1}{3} - \frac{6}{3} = -\frac{5}{3}$$
2. **Relation with coefficients:**
Comparing $3x^2 + 5x - 2$ with $ax^2 + bx + c$, we have $a = 3$, $b = 5$, and $c = -2$.
$$\text{Sum of zeroes} = -\frac{b}{a} = -\frac{5}{3}$$
Since both values are equal ($-\frac{5}{3} = -\frac{5}{3}$), the relation between the sum of zeroes and the coefficients is verified.
Q31. If one zero of the quadratic polynomial $x^2 - 3x + k$ is 2, find the value of $k$.
Answer: Value of $k$: $2$
Justification & Calculation:
Let the given polynomial be $p(x) = x^2 - 3x + k$.
Since $2$ is a zero of the polynomial, substituting $x = 2$ must make the polynomial equal to zero ($p(2) = 0$):
$$(2)^2 - 3(2) + k = 0$$
$$4 - 6 + k = 0$$
$$-2 + k = 0$$
$$k = 2$$
Therefore, the value of $k$ is $2$.
Q32. Find the value of $k$ if the sum of the zeroes of the polynomial $x^2 - (k+6)x + 2(2k-1)$ is equal to half of their product.
Answer: Value of $k$: $7$
Justification & Calculation:
Let the given polynomial be $p(x) = x^2 - (k+6)x + 2(2k-1)$.
Comparing this with the standard quadratic polynomial $ax^2 + bx + c$, we get:
$$a = 1, \quad b = -(k+6), \quad c = 2(2k-1)$$
Let $\alpha$ and $\beta$ be the zeroes of the polynomial. Using Vieta's formulas:
$$\text{Sum of zeroes } (\alpha + \beta) = -\frac{b}{a} = \frac{k+6}{1} = k + 6$$
$$\text{Product of zeroes } (\alpha\beta) = \frac{c}{a} = \frac{2(2k-1)}{1} = 4k - 2$$
According to the given condition, the sum of the zeroes is equal to half of their product:
$$\alpha + \beta = \frac{1}{2}\alpha\beta$$
$$k + 6 = \frac{1}{2}(4k - 2)$$
$$k + 6 = 2k - 1$$
$$6 + 1 = 2k - k$$
$$k = 7$$
Therefore, the value of $k$ is $7$.
Q33. If $\alpha$ and $\beta$ are the zeroes of the polynomial $2x^2 + 5x + k$ such that $\alpha^2 + \beta^2 + \alpha\beta = \frac{21}{4}$, find the value of $k$.
Answer: Value of $k$: $2$
Justification & Calculation:
Let the given polynomial be $p(x) = 2x^2 + 5x + k$.
Comparing this with $ax^2 + bx + c$, we get:
$$a = 2, \quad b = 5, \quad c = k$$
Using Vieta's formulas for the zeroes $\alpha$ and $\beta$:
$$\alpha + \beta = -\frac{b}{a} = -\frac{5}{2}$$
$$\alpha\beta = \frac{c}{a} = \frac{k}{2}$$
We are given the relation:
$$\alpha^2 + \beta^2 + \alpha\beta = \frac{21}{4}$$
Rewriting $\alpha^2 + \beta^2$ using the identity $(\alpha + \beta)^2 = \alpha^2 + \beta^2 + 2\alpha\beta$:
$$(\alpha + \beta)^2 - 2\alpha\beta + \alpha\beta = \frac{21}{4}$$
$$(\alpha + \beta)^2 - \alpha\beta = \frac{21}{4}$$
Substitute the values of $(\alpha + \beta)$ and $\alpha\beta$ into the equation:
$$\left(-\frac{5}{2}\right)^2 - \frac{k}{2} = \frac{21}{4}$$
$$\frac{25}{4} - \frac{k}{2} = \frac{21}{4}$$
$$\frac{25}{4} - \frac{21}{4} = \frac{k}{2}$$
$$\frac{4}{4} = \frac{k}{2}$$
$$1 = \frac{k}{2}$$
$$k = 2$$
Therefore, the value of $k$ is $2$.
Q34. If one zero of the quadratic polynomial $4x^2 - 8kx - 9$ is the negative of the other, find the value of $k$.
Answer: Value of $k$: $0$
Justification & Calculation:
Let the given quadratic polynomial be $p(x) = 4x^2 - 8kx - 9$.
Comparing this with the standard form $ax^2 + bx + c$, we get:
$$a = 4, \quad b = -8k, \quad c = -9$$
Let the zeroes of the polynomial be $\alpha$ and $\beta$. According to the problem, one zero is the negative of the other, so let:
$$\alpha = m \quad \text{and} \quad \beta = -m$$
Using Vieta's formulas:
$$\text{Sum of zeroes } (\alpha + \beta) = -\frac{b}{a}$$
$$m + (-m) = -\frac{-8k}{4}$$
$$0 = \frac{8k}{4}$$
$$2k = 0$$
$$k = 0$$
Therefore, the value of $k$ is $0$.
Q35. Find the value of $a$ if the product of the zeroes of the polynomial $ax^2 - 6x - 6$ is 4.
Answer: Value of $a$: $-\frac{3}{2}$
Justification & Calculation:
Let the given quadratic polynomial be $p(x) = ax^2 - 6x - 6$.
Comparing this with the standard form $Ax^2 + Bx + C$, we get:
$$A = a, \quad B = -6, \quad C = -6$$
Using Vieta's formulas, the product of the zeroes for a quadratic polynomial is given by $\frac{C}{A}$:
$$\text{Product of zeroes} = \frac{-6}{a}$$
According to the problem, the product of the zeroes is $4$:
$$\frac{-6}{a} = 4$$
$$4a = -6$$
$$a = -\frac{6}{4}$$
$$a = -\frac{3}{2}$$
Therefore, the value of $a$ is $-\frac{3}{2}$.
Q36. If $\alpha$ and $\beta$ are the zeroes of the polynomial $x^2 - px + q$, find the value of $\frac{1}{\alpha} + \frac{1}{\beta}$.
Answer: Value of $\frac{1}{\alpha} + \frac{1}{\beta}$: $\frac{p}{q}$
Justification & Calculation:
Let the given polynomial be $p(x) = x^2 - px + q$.
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 $\beta$:
$$\text{Sum of zeroes } (\alpha + \beta) = -\frac{b}{a} = -\frac{-p}{1} = p$$
$$\text{Product of zeroes } (\alpha\beta) = \frac{c}{a} = \frac{q}{1} = q$$
Now, simplify the required expression:
$$\frac{1}{\alpha} + \frac{1}{\beta} = \frac{\alpha + \beta}{\alpha\beta}$$
Substitute the values of $(\alpha + \beta)$ and $\alpha\beta$ into the expression:
$$\frac{1}{\alpha} + \frac{1}{\beta} = \frac{p}{q}$$
Therefore, the value of $\frac{1}{\alpha} + \frac{1}{\beta}$ is $\frac{p}{q}$.
Q37. If $\alpha$ and $\beta$ are zeroes of $x^2 - x - 4$, find the value of $\alpha^2\beta + \alpha\beta^2$.
Answer: Value of $\alpha^2\beta + \alpha\beta^2$: $4$
Justification & Calculation:
Let the given polynomial be $p(x) = x^2 - x - 4$.
Comparing this with the standard quadratic form $ax^2 + bx + c$, we get:
$$a = 1, \quad b = -1, \quad c = -4$$
Using Vieta's formulas for the zeroes $\alpha$ and $\beta$:
$$\alpha + \beta = -\frac{b}{a} = -\frac{-1}{1} = 1$$
$$\alpha\beta = \frac{c}{a} = \frac{-4}{1} = -4$$
Now, factorize the given expression $\alpha^2\beta + \alpha\beta^2$:
$$\alpha^2\beta + \alpha\beta^2 = \alpha\beta(\alpha + \beta)$$
Substitute the values of $(\alpha + \beta)$ and $\alpha\beta$ into the factored expression:
$$\alpha^2\beta + \alpha\beta^2 = (-4)(1) = -4$$
Wait, let me double check the calculation:
$\alpha\beta = -4$, $\alpha + \beta = 1$.
Product = $-4 \times 1 = -4$.
Let me re-read the values: $x^2 - x - 4$. $b = -1$, so $-\frac{b}{a} = 1$. $c = -4$, so $\frac{c}{a} = -4$.
Expression: $\alpha^2\beta + \alpha\beta^2 = \alpha\beta(\alpha + \beta) = (-4)(1) = -4$.
Ah, let me format the answer cleanly as $-4$.
Value of $\alpha^2\beta + \alpha\beta^2$: $-4$
Justification & Calculation:
Let the given polynomial be $p(x) = x^2 - x - 4$.
Comparing this with the standard quadratic form $ax^2 + bx + c$, we get:
$$a = 1, \quad b = -1, \quad c = -4$$
Using Vieta's formulas for the zeroes $\alpha$ and $\beta$:
$$\alpha + \beta = -\frac{b}{a} = -\frac{-1}{1} = 1$$
$$\alpha\beta = \frac{c}{a} = \frac{-4}{1} = -4$$
Now, factorize the given expression:
$$\alpha^2\beta + \alpha\beta^2 = \alpha\beta(\alpha + \beta)$$
Substitute the values:
$$\alpha^2\beta + \alpha\beta^2 = (-4)(1) = -4$$
Therefore, the value of $\alpha^2\beta + \alpha\beta^2$ is $-4$.
Q38. If the zeroes of the polynomial $x^2 - kx + 6$ are in the ratio $2:3$, find the positive value of $k$.
Answer: Positive value of $k$: $5\sqrt{6}$ (or approximately $12.25$, depending on whether it's $5$ or $5\sqrt{6}$—let's compute carefully: ratio $2:3$, let zeroes be $2m$ and $3m$. Sum = $5m = k$, Product = $6m^2 = 6 \implies m^2 = 1 \implies m = \pm 1$. If $m = 1$, sum = $5$, product = $6 \implies k = 5$. Let me re-verify: $x^2 - 5x + 6$, zeroes are $2$ and $3$, ratio is $2:3$, sum is $5 \implies k = 5$).
Justification & Calculation:
Let the given polynomial be $p(x) = x^2 - kx + 6$.
Comparing this with the standard form $ax^2 + bx + c$, we get:
$$a = 1, \quad b = -k, \quad c = 6$$
Since the zeroes are in the ratio $2:3$, let the zeroes be $2m$ and $3m$ for some non-zero constant $m$.
Using Vieta's formulas:
$$\text{Sum of zeroes } (2m + 3m) = -\frac{-k}{1} \implies 5m = k$$
$$\text{Product of zeroes } (2m \cdot 3m) = \frac{6}{1} \implies 6m^2 = 6$$
Solving for $m^2$:
$$m^2 = \frac{6}{6} = 1$$
$$m = \pm 1$$
Since we need the positive value of $k$, let's consider both cases for $m$:
* If $m = 1$: $k = 5(1) = 5$
* If $m = -1$: $k = 5(-1) = -5$
Choosing the positive value, we get $k = 5$.
Therefore, the positive value of $k$ is $5$.
Q39. If $\alpha$ and $\beta$ are zeroes of the polynomial $p(x) = 2x^2 + 3x - 1$, find the value of $\alpha^2 + \beta^2$.
Answer: Value of $\alpha^2 + \beta^2$: $\frac{13}{4}$
Justification & Calculation:
Let the given polynomial be $p(x) = 2x^2 + 3x - 1$.
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 $\beta$:
$$\alpha + \beta = -\frac{b}{a} = -\frac{3}{2}$$
$$\alpha\beta = \frac{c}{a} = \frac{-1}{2}$$
Using the algebraic identity for the sum of squares:
$$\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta$$
Substitute the values of $(\alpha + \beta)$ and $\alpha\beta$ into the identity:
$$\alpha^2 + \beta^2 = \left(-\frac{3}{2}\right)^2 - 2\left(-\frac{1}{2}\right)$$
$$\alpha^2 + \beta^2 = \frac{9}{4} + 1$$
$$\alpha^2 + \beta^2 = \frac{9}{4} + \frac{4}{4} = \frac{13}{4}$$
Therefore, the value of $\alpha^2 + \beta^2$ is $\frac{13}{4}$.
Q40. If one zero of the polynomial $(a^2 + 9)x^2 + 13x + 6a$ is the reciprocal of the other, find the value of $a$.
Answer: Value of $a$: $\frac{3}{2}$
Justification & Calculation:
Let the given polynomial be $p(x) = (a^2 + 9)x^2 + 13x + 6a$.
Comparing this with the standard quadratic form $Ax^2 + Bx + C$, we get:
$$A = a^2 + 9, \quad B = 13, \quad C = 6a$$
Let the zeroes of the polynomial be $\alpha$ and $\frac{1}{\alpha}$ since one zero is the reciprocal of the other.
Using Vieta's formulas, the product of the zeroes is given by $\frac{C}{A}$:
$$\alpha \cdot \frac{1}{\alpha} = \frac{6a}{a^2 + 9}$$
$$1 = \frac{6a}{a^2 + 9}$$
Cross-multiply to solve for $a$:
$$a^2 + 9 = 6a$$
$$a^2 - 6a + 9 = 0$$
This is a perfect square trinomial:
$$(a - 3)^2 = 0$$
$$a - 3 = 0$$
$$a = 3$$
Wait, let me double check the calculation:
If $a = 3$, $C = 6(3) = 18$, $A = 3^2 + 9 = 18$, product $\frac{18}{18} = 1$. Correct.
Wait, let me re-verify: $6a / (a^2+9) = 1 \implies a^2 - 6a + 9 = 0 \implies (a-3)^2 = 0 \implies a = 3$.
Let me correct the final value in the answer text to $3$ instead of $\frac{3}{2}$.
Value of $a$: $3$
Justification & Calculation:
Let the given polynomial be $p(x) = (a^2 + 9)x^2 + 13x + 6a$.
Comparing this with the standard quadratic form $Ax^2 + Bx + C$, we get:
$$A = a^2 + 9, \quad B = 13, \quad C = 6a$$
Let the zeroes be $\alpha$ and $\frac{1}{\alpha}$. Using the product of zeroes:
$$\alpha \cdot \frac{1}{\alpha} = \frac{6a}{a^2 + 9}$$
$$1 = \frac{6a}{a^2 + 9}$$
$$a^2 + 9 = 6a$$
$$a^2 - 6a + 9 = 0$$
$$(a - 3)^2 = 0$$
$$a = 3$$
Therefore, the value of $a$ is $3$.
Q41. If the sum of the zeroes of the quadratic polynomial $ky^2 + 2y + 3k$ 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 $p(y) = ky^2 + 2y + 3k$.
Comparing this with the standard form $ay^2 + by + c$, we get:
$$a = k, \quad b = 2, \quad c = 3k$$
Using Vieta's formulas for the zeroes:
$$\text{Sum of zeroes} = -\frac{b}{a} = -\frac{2}{k}$$
$$\text{Product of zeroes} = \frac{c}{a} = \frac{3k}{k} = 3$$
According to the given condition, the sum of the zeroes is equal to their product:
$$-\frac{2}{k} = 3$$
$$-2 = 3k$$
$$k = -\frac{2}{3}$$
Therefore, the value of $k$ is $-\frac{2}{3}$.
Q42. If $\alpha, \beta$ are the zeroes of the polynomial $f(x) = x^2 - 5x + k$ such that $\alpha - \beta = 1$, find the value of $k$.
Answer: Value of $k$: $6$
Justification & Calculation:
Let the given polynomial be $f(x) = x^2 - 5x + k$.
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 $\beta$:
$$\alpha + \beta = -\frac{b}{a} = -\frac{-5}{1} = 5$$
$$\alpha\beta = \frac{c}{a} = \frac{k}{1} = k$$
We are given the condition:
$$\alpha - \beta = 1$$
We know the algebraic identity:
$$(\alpha - \beta)^2 = (\alpha + \beta)^2 - 4\alpha\beta$$
Substitute the known values into the identity:
$$(1)^2 = (5)^2 - 4(k)$$
$$1 = 25 - 4k$$
$$4k = 25 - 1$$
$$4k = 24$$
$$k = 6$$
Therefore, the value of $k$ is $6$.
Q43. Find a quadratic polynomial, the sum and product of whose zeroes are $-3$ and $2$ respectively.
Answer: Quadratic polynomial: $x^2 + 3x + 2$ (or any polynomial of the form $k(x^2 + 3x + 2)$ where $k$ is a non-zero real number)
Justification & Calculation:
Let the sum of the zeroes be $S$ and the product of the zeroes be $P$.
From the given problem:
$$S = -3 \quad \text{and} \quad P = 2$$
A quadratic polynomial can be expressed in terms of its sum and product of zeroes as:
$$p(x) = k \left[ x^2 - (\text{Sum of zeroes})x + (\text{Product of zeroes}) \right]$$
where $k$ is a non-zero real constant.
Substitute the given values into the formula:
$$p(x) = k \left[ x^2 - (-3)x + 2 \right]$$
$$p(x) = k \left( x^2 + 3x + 2 \right)$$
For the simplest polynomial (taking $k = 1$):
$$p(x) = x^2 + 3x + 2$$
Therefore, the required quadratic polynomial is $x^2 + 3x + 2$.
Q44. Form a quadratic polynomial whose zeroes are $2$ and $-6$.
Answer: Quadratic polynomial: $x^2 + 4x - 12$ (or any polynomial of the form $k(x^2 + 4x - 12)$ where $k$ is a non-zero real number)
Justification & Calculation:
Let the given zeroes be $\alpha = 2$ and $\beta = -6$.
1. **Find the sum of the zeroes ($S$):**
$$S = \alpha + \beta = 2 + (-6) = -4$$
2. **Find the product of the zeroes ($P$):**
$$P = \alpha\beta = (2)(-6) = -12$$
A quadratic polynomial can be formed using the formula:
$$p(x) = k \left[ x^2 - Sx + P \right]$$
where $k$ is a non-zero real constant.
Substitute the values of $S$ and $P$ into the formula:
$$p(x) = k \left[ x^2 - (-4)x + (-12) \right]$$
$$p(x) = k \left( x^2 + 4x - 12 \right)$$
For the simplest polynomial (taking $k = 1$):
$$p(x) = x^2 + 4x - 12$$
Therefore, the required quadratic polynomial is $x^2 + 4x - 12$.
Q45. Find a quadratic polynomial whose sum of zeroes is $\frac{1}{4}$ and product of zeroes is $-1$.
Answer: Quadratic polynomial: $4x^2 - x - 4$ (or any polynomial of the form $k(4x^2 - x - 4)$ where $k$ is a non-zero real number)
Justification & Calculation:
Let the sum of the zeroes be $S$ and the product of the zeroes be $P$.
From the given problem:
$$S = \frac{1}{4} \quad \text{and} \quad P = -1$$
A quadratic polynomial can be expressed in terms of its sum and product of zeroes as:
$$p(x) = k \left[ x^2 - (\text{Sum of zeroes})x + (\text{Product of zeroes}) \right]$$
where $k$ is a non-zero real constant.
Substitute the given values into the formula:
$$p(x) = k \left[ x^2 - \frac{1}{4}x + (-1) \right]$$
$$p(x) = k \left( x^2 - \frac{1}{4}x - 1 \right)$$
To clear the fraction and obtain the simplest polynomial with integer coefficients, we can choose $k = 4$:
$$p(x) = 4\left( x^2 - \frac{1}{4}x - 1 \right)$$
$$p(x) = 4x^2 - x - 4$$
Therefore, the required quadratic polynomial is $4x^2 - x - 4$.
Q46. Find a quadratic polynomial whose zeroes are $3 + \sqrt{2}$ and $3 - \sqrt{2}$.
Answer: Quadratic polynomial: $x^2 - 6x + 7$ (or any polynomial of the form $k(x^2 - 6x + 7)$ where $k$ is a non-zero real number)
Justification & Calculation:
Let the given zeroes be $\alpha = 3 + \sqrt{2}$ and $\beta = 3 - \sqrt{2}$.
1. **Find the sum of the zeroes ($S$):**
$$S = \alpha + \beta = (3 + \sqrt{2}) + (3 - \sqrt{2}) = 3 + 3 = 6$$
2. **Find the product of the zeroes ($P$):**
$$P = \alpha\beta = (3 + \sqrt{2})(3 - \sqrt{2})$$
Using the algebraic identity $(a + b)(a - b) = a^2 - b^2$:
$$P = (3)^2 - (\sqrt{2})^2 = 9 - 2 = 7$$
A quadratic polynomial can be formed using the formula:
$$p(x) = k \left[ x^2 - Sx + P \right]$$
where $k$ is a non-zero real constant.
Substitute the values of $S$ and $P$ into the formula:
$$p(x) = k \left( x^2 - 6x + 7 \right)$$
For the simplest polynomial (taking $k = 1$):
$$p(x) = x^2 - 6x + 7$$
Therefore, the required quadratic polynomial is $x^2 - 6x + 7$.
Q47. Write the quadratic polynomial whose sum and product of zeroes are $\sqrt{2}$ and $\frac{1}{3}$ respectively.
Answer: Quadratic polynomial: $3x^2 - 3\sqrt{2}x + 1$ (or any polynomial of the form $k(3x^2 - 3\sqrt{2}x + 1)$ where $k$ is a non-zero real number)
Justification & Calculation:
Let the sum of the zeroes be $S$ and the product of the zeroes be $P$.
From the given problem:
$$S = \sqrt{2} \quad \text{and} \quad P = \frac{1}{3}$$
A quadratic polynomial can be expressed in terms of its sum and product of zeroes as:
$$p(x) = k \left[ x^2 - (\text{Sum of zeroes})x + (\text{Product of zeroes}) \right]$$
where $k$ is a non-zero real constant.
Substitute the given values into the formula:
$$p(x) = k \left[ x^2 - \sqrt{2}x + \frac{1}{3} \right]$$
To clear the fraction and obtain the simplest polynomial with integer coefficients, we can choose $k = 3$:
$$p(x) = 3 \left( x^2 - \sqrt{2}x + \frac{1}{3} \right)$$
$$p(x) = 3x^2 - 3\sqrt{2}x + 1$$
Therefore, the required quadratic polynomial is $3x^2 - 3\sqrt{2}x + 1$.
Q48. Form a quadratic polynomial whose zeroes are the reciprocals of the zeroes of the polynomial $ax^2 + bx + c$.
Answer: Quadratic polynomial: $cx^2 + bx + a$ (or any polynomial of the form $k(cx^2 + bx + a)$ where $k$ is a non-zero real number)
Justification & Calculation:
Let $\alpha$ and $\beta$ be the zeroes of the given polynomial $p(x) = ax^2 + bx + c$.
Using Vieta's formulas:
$$\alpha + \beta = -\frac{b}{a}$$
$$\alpha\beta = \frac{c}{a}$$
We need to form a quadratic polynomial whose zeroes are the reciprocals of $\alpha$ and $\beta$, which are $\frac{1}{\alpha}$ and $\frac{1}{\beta}$.
1. **Find the sum of the new zeroes ($S'$):**
$$S' = \frac{1}{\alpha} + \frac{1}{\beta} = \frac{\alpha + \beta}{\alpha\beta}$$
Substitute the values from Vieta's formulas:
$$S' = \frac{-\frac{b}{a}}{\frac{c}{a}} = -\frac{b}{c}$$
2. **Find the product of the new zeroes ($P'$):**
$$P' = \frac{1}{\alpha} \cdot \frac{1}{\beta} = \frac{1}{\alpha\beta}$$
Substitute the value from Vieta's formulas:
$$P' = \frac{1}{\frac{c}{a}} = \frac{a}{c}$$
A quadratic polynomial can be formed using the formula:
$$p'(x) = k \left[ x^2 - S'x + P' \right]$$
where $k$ is a non-zero real constant.
Substitute $S'$ and $P'$ into the formula:
$$p'(x) = k \left[ x^2 - \left(-\frac{b}{c}\right)x + \frac{a}{c} \right]$$
$$p'(x) = k \left[ x^2 + \frac{b}{c}x + \frac{a}{c} \right]$$
To clear the fraction, we can choose $k = c$:
$$p'(x) = c \left( x^2 + \frac{b}{c}x + \frac{a}{c} \right)$$
$$p'(x) = cx^2 + bx + a$$
Therefore, the required quadratic polynomial is $cx^2 + bx + a$.
Q49. Find a quadratic polynomial whose zeroes are $\frac{2}{3}$ and $-\frac{1}{4}$.
Answer: Quadratic polynomial: $12x^2 - 5x - 2$ (or any polynomial of the form $k(12x^2 - 5x - 2)$ where $k$ is a non-zero real number)
Justification & Calculation:
Let the given zeroes be $\alpha = \frac{2}{3}$ and $\beta = -\frac{1}{4}$.
1. **Find the sum of the zeroes ($S$):**
$$S = \alpha + \beta = \frac{2}{3} + \left(-\frac{1}{4}\right) = \frac{8 - 3}{12} = \frac{5}{12}$$
2. **Find the product of the zeroes ($P$):**
$$P = \alpha\beta = \left(\frac{2}{3}\right)\left(-\frac{1}{4}\right) = -\frac{2}{12} = -\frac{1}{6}$$
A quadratic polynomial can be formed using the formula:
$$p(x) = k \left[ x^2 - Sx + P \right]$$
where $k$ is a non-zero real constant.
Substitute the values of $S$ and $P$ into the formula:
$$p(x) = k \left[ x^2 - \frac{5}{12}x - \frac{1}{6} \right]$$
To clear the fractions and obtain the simplest polynomial with integer coefficients, we can choose $k = 12$:
$$p(x) = 12 \left( x^2 - \frac{5}{12}x - \frac{1}{6} \right)$$
$$p(x) = 12x^2 - 5x - 2$$
Therefore, the required quadratic polynomial is $12x^2 - 5x - 2$.
Q50. Find a quadratic polynomial the sum of whose zeroes is 0 and one zero is 5.
Answer: Quadratic polynomial: $x^2 - 25$ (or any polynomial of the form $k(x^2 - 25)$ where $k$ is a non-zero real number)
Justification & Calculation:
Let the zeroes of the quadratic polynomial be $\alpha$ and $\beta$.
We are given:
* Sum of the zeroes: $\alpha + \beta = 0$
* One zero: $\alpha = 5$
Substitute $\alpha = 5$ into the sum of zeroes equation to find the other zero ($\beta$):
$$5 + \beta = 0$$
$$\beta = -5$$
Now, find the product of the zeroes ($P$):
$$P = \alpha\beta = (5)(-5) = -25$$
We are also given the sum of the zeroes ($S$):
$$S = 0$$
A quadratic polynomial can be formed using the formula:
$$p(x) = k \left[ x^2 - Sx + P \right]$$
where $k$ is a non-zero real constant.
Substitute $S = 0$ and $P = -25$ into the formula:
$$p(x) = k \left[ x^2 - (0)x + (-25) \right]$$
$$p(x) = k \left( x^2 - 25 \right)$$
For the simplest polynomial (taking $k = 1$):
$$p(x) = x^2 - 25$$
Therefore, the required quadratic polynomial is $x^2 - 25$.
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.
