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

Answer

The given quadratic equation is $4 s^2-4 s+1$
$=(2 s)^2-2(2 s) 1+1^2$
As, we know $(a-b)^2=a^2-2 a b+b^2$, the above equation can be written as
$=(2 s-1)^2$
The value of $4 s^2-4 s+1$ is zero when $2 s-1=0$, when, $s=\frac{1}{2}, \frac{1}{2}$
Therefore, the zeroes of $4 s^2-4 s+1$ 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

Find the zeroes of the quadratic polynomial $6 x^2-3-7 x$

Answer

$\begin{aligned} 6 x^2-3-7 x & =6 x^2-7 x-3
\\ & =6 x^2-9 x+2 x-3
\\ & =3 x(2 x-3)+1(2 x-3)
\\ & =(2 x-3)(3 x+1)\end{aligned}$
So, $6 x^2-3-7 x=0$ when $2 x-3=0$ or $3 x+1=0$
i.e., when $x=\frac{3}{2}$ or $x=-\frac{1}{3}$.
Hence, the zeroes of the polynomial $6 x^2-3-7 x$ are $\frac{3}{2}$ and $-\frac{1}{3}$.

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Question 102 Marks

Find a quadratic polynomial, the sum and product of whose zeroes are -3 and 2 respectively.

Answer

The standard form of quadratic polynomial be given as: ax² + bx + c, and its zeroes will be α and β.
We have
$\alpha+\beta=-3=\frac{-b}{a}$
and $\alpha \beta=2=\frac{c}{a}$
If a = 1, then b = 3 and c = 2. So, one quadratic polynomial which fits the given conditions is x² + 3x + 2

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Question 112 Marks

If one root of quadratic polynomial $6 x^2+37 x-( P -2)$ is inverse of the other root, then find the value of P.

Answer

For a quadratic polynomial $a x^2+b x+c$, the product of the roots is given by the formula:
Product of roots $=\frac{c}{a}$
In the given polynomial $6 x^2+37 x-(P-2):$
$a=6$
$c=-(P-2)=2-P$
Since the product of the roots is 1, we can set up the equation:
$\begin{array}{l}\frac{2-P}{6}=1 \\ 2-P=6 \\ -P=6-2 \\ -P=4 \\ P=-4\end{array}$
The value of P is -4.

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Question 122 Marks

Find the roots of the quadratic equation $x^2-x-20=0$

Answer

The roots of the quadratic equation $x^2-x-20=0$ are 5 and -4.
Factor the quadratic equation by finding two numbers that multiply to −20 and add up to −1. These numbers are 5 and −4.
Rewrite the equation using these numbers:
$(x-5)(x+4)=0$
Set each factor equal to zero to solve for $x:$
$\begin{array}{l}x-5=0 \Longrightarrow x=5 \\ x+4=0 \Longrightarrow x=-4\end{array}$

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