2 Marks Questions

Question 12 Marks

Find a quadratic polynomial, the sum and product of whose zeroes are $\sqrt { 2 } , \frac { 1 } { 3 }$ respectively.

Answer

Let the polynomial be $ax^2+ bx + c.$
and its zeroes be $\alpha$ and $\beta$.
Then, $\alpha + \beta = \sqrt { 2 } = - \frac { b } { a } \text { and } \alpha \beta = \frac { 1 } { 3 } = \frac { c } { a }$
If $a = 3$, then $b = - 3 \sqrt { 2 }$ and $c = 1$.
So, one quadratic polynomial which fits the given conditions is $3 x ^ { 2 } - 3 \sqrt { 2 } x + 1$.
It is given that $\alpha + \beta = \sqrt { 2 }$ and $\alpha \beta = \frac { 1 } { 3 }$
Now, standard form of quadratic polynomial is given by $x ^ { 2 } - ( \alpha + \beta ) x + \alpha \beta$
$= x ^ { 2 } - ( \alpha + \beta ) x + \alpha \beta$
$= x ^ { 2 } - \sqrt { 2 } x + \frac { 1 } { 3 }$
$= \frac { 1 } { 3 } \left( 3 x ^ { 2 } - 3 \sqrt { 2 } x + 1 \right)$
Hence the required quadratic polynomial is $3 x ^ { 2 } - 3 \sqrt { 2 } x + 1$

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Question 22 Marks

Find a quadratic polynomial, the sum and product of whose zeroes are $\frac { 1 } { 4 } , - 1$ respectively.

Answer

Let the required polynomial be $ax^2+ bx + c$
and let its zeroes be $\alpha$ and $\beta$
Then, $\alpha + \beta = \frac { 1 } { 4 } = - \frac { b } { a } \text { and } \alpha \beta = - 1 = \frac { c } { a }$
If $a = 4$, then $b = -1$ and $c = -4$
So, one quadratic polynomial which satisfies the given conditions is $4x^2- x - 4$
Or
If $\alpha$ and $\beta$ zeroes of the polynomials then standard quadratic polynomial is given by
$x ^ { 2 } - ( \alpha + \beta ) x + \alpha \beta$, where $\alpha + \beta = \frac14 ~and~ \alpha\beta=-1 $ [Given]
Now, we have,
$ x ^ { 2 } - ( \alpha + \beta ) x + \alpha \beta$
$= x ^ { 2 } - \left( \frac { 1 } { 4 } \right) x + ( - 1 )$
$= \frac { 1 } { 4 } \left( 4 x ^ { 2 } - x - 4 \right)$
Required polynomial is $4x^2- x - 4$

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Question 32 Marks

Find the zeroes of quadratic polynomial $t^2– 15$ and verify the relationship between the zeroes and their coefficients.

Answer

We have quadratic polynomial as $t^2– 15$
$= t^2– (\sqrt{15})^2$
$= (t - \sqrt{15})(t + \sqrt{15}) [$As,$x^2- y^2= (x - y)(x + y)]$
The value of $t^2− 15$ is zero when $(t -\sqrt{15}) = 0$ or $(t + \sqrt{15}) = 0,$
i.e., when $t = \sqrt{15}$ or $t = -\sqrt{15}$
therefore, the zeroes of $t^2− 15$ are $\sqrt{15}$ and $- \sqrt{15}$.
Sum of zeroes = $\sqrt{15}+(-\sqrt{15})=0=\frac{-0}{1}=\frac{-(\text {coefficient of } t)}{\text {coefficient of } t^{2}}$
Product of zeroes = $(\sqrt{15})(-\sqrt{15})=-15=\frac{-15}{1}=\frac{\text { constant term }}{\text { coefficient of } t^{2}}$
Hence verified.

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Question 42 Marks

Find the zeroes of quadratic polynomial $4u^2+ 8u$ and verify the relationship between the zeroes and their coefficients.

Answer

The quadratic equation is given as: $4u^2+ 8u$
it can be written in the standard form as:
$= 4u^2+ 8u + 0$
$= 4u (u + 2)$
The value of $4u^2+ 8u$ is zero when $4u = 0$ or $u + 2 = 0,$
$i.e., u = 0$ or $u = −2$
Therefore, the zeroes of $4u^2+ 8u$ are $0$ and $−2$
Sum of zeroes = $0+(-2)=-2=\frac{-(8)}{4}=\frac{-(\text { coefficient of } u)}{\text { coefficient of } u^{2}}$
Product of zeroes = $0 \times(-2)=0=\frac{0}{4}=\frac{\text { constant torm }}{\text { coefficient of } u^{2}}$
Hence verified

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Question 52 Marks

Find the zeroes of quadratic polynomial $6x^2- 3 - 7x$ and verify the relationship between the zeroes and their coefficients.

Answer

We have given the quadratic equation as: $6x^2- 3 - 7x$
First of all we will write it into standard form as: $6x^2- 7x - 3$
(Now we will factorize $7$ such that the product of the factors is equal to $- 18$ and the sum is equal to $- 7)$
It can be written as
$= 6x^2+ 2x - 9x - 3$
$= 2x(3x + 1) - 3(3x + 1)$
$= (3x + 1)(2x - 3)$
The value of $6x^2- 3 - 7x $ is zero when $3x + 1 = 0$ or $2x − 3 = 0,$
i.e. $X=\frac{-1}{3} \text { or } \frac{3}{2}$
Therefore, the zeroes of $6x^2− 3 − 7x $ are $\frac{-1}{3} \text { and } \frac{3}{2}$
Sum of zeroes = $\frac{-1}{3}+\frac{3}{2}=\frac{7}{6}=\frac{-(-7)}{6}=\frac{-(\text { coefficient of } x)}{\text { coefficient of } x^{2}}$
Product of zeroes =$\frac{-1}{3} \times \frac{3}{2}=\frac{-1}{2}=\frac{-3}{6}=\frac{\text { constant term }}{\text { coefficient of } x^{2}}$
Hence, verified

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Question 62 Marks

Find the zeroes of quadratic polynomial $4s^2- 4s + 1$ and verify the relationship between the zeroes and their coefficients.

Answer

The given quadratic equation is $4s^2- 4s + 1$
$= (2s)^2- 2(2s)1 + 1^2$
As, we know $(a - b)^2= a^2- 2ab + b^2$, the above equation can be written as
$= (2s - 1)^2$
The value of $4s^2− 4s + 1$ is zero when $2s − 1 = 0,$ when, $s = \frac{1}{2}$, $\frac{1}{2}$
Therefore, the zeroes of $4s^2− 4s + 1$ are $\frac{1}{2}$ and $\frac{1}{2}$
Sum of zeroes = $\frac{1}{2}+\frac{1}{2}=1=\frac{-(-4)}{4}=\frac{-(\text {coefficient of } s)}{\text {coefficient of } s^{2}}$
Product of zeroes = $\frac{1}{2} \times \frac{1}{2}=\frac{1}{4}=\frac{\text { constant term }}{\text { coefficient of } s^{2}}$
Hence Verified.

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Question 72 Marks

Find the zeroes of the polynomial $x^2– 3$ and verify the relationship between the zeroes and the coefficients.

Answer

Given quadratic equation: $x^2- 3$
Recall the identity $a^2– b^2= (a – b)(a + b).$ Using it,
we can write:
$x^{2}-3=(x-\sqrt{3})(x+\sqrt{3})$
So, the value of $x^2– 3$ is zero when $x = \sqrt{3}$ or $ x = -\sqrt{3}$
Therefore, the zeroes of $x^{2}-3 \text { are } \sqrt{3} \text { and }-\sqrt{3}$
Now, the sum of zeroes $= \sqrt{3}-\sqrt{3}=0=\frac{-(\text { Coefficient of } x)}{\text { Coefficient of } x^{2}}$
and the product of zeroes $=(\sqrt{3})(-\sqrt{3})=-3=\frac{-3}{1}=\frac{\text { Constant term }}{\text { Coefficient of } x^{2}}$

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Question 82 Marks

Find the zeroes of the quadratic polynomial $x^2+ 7x + 10,$ and verify the relationship between the zeroes and the coefficients.

Answer

We have,
$x^2+ 7x + 10 = (x + 2)(x + 5)$
So, the value of $x^2+ 7x + 10$ is zero when $x + 2 = 0$ or $x + 5 = 0,$ i.e., when $x = – 2$ or $x = –5.$
Therefore, the zeroes of $x^2+ 7x + 10$ are $-2$ and $-5.$
Now,
sum of zeroes = $-2+(-5)=-(7)=\frac{-(7)}{1}=\frac{-(\text { Coefficient of } x)}{\text { Coefficient of } x^{2}}$
product of zeroes = $(-2) \times(-5)=10=\frac{10}{1}=\frac{\text { Constant term }}{\text { coefficient of } x^{2}}$

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Question 92 Marks

If the zeros of polynomial are $2+\sqrt{3}$ and $2-\sqrt{3}$ then find the polynomial.

Answer

Let the zeros of the polynomial be $\alpha=2+\sqrt{3}$ and $\beta=2-\sqrt{3}$.
The general form of a quadratic polynomial is given by $x^2-(\alpha+\beta) x+\alpha \beta$.
Sum of the Zeros $(\alpha+\beta):$
$\begin{array}{l}\alpha+\beta=(2+\sqrt{3})+(2-\sqrt{3}) \\ \alpha+\beta=4\end{array}$
Product of the Zeros $(\alpha \beta):$
$\begin{array}{l}\alpha \beta=(2+\sqrt{3})(2-\sqrt{3}) \\ \alpha \beta=(2)^2-(\sqrt{3})^2=4-3=1\end{array}$
Substituting these values into the polynomial form:
$x^2-(4) x+1$
Therefore, the required polynomial is $x^2-4 x+1$

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