3 Marks Question

Question 13 Marks

Find a quadratic polynomial of the given numbers as the sum and product of its zeroes respectively. $- \frac { 1 } { 4 } , \frac { 1 } { 4 }$

Answer

Let the polynomial be $ax^2+ bx + c,$
and its zeroes be $\alpha $ and $\beta $.
Then, $\alpha + \beta = - \frac { 1 } { 4 } = - \frac { b } { a } \text { and } \alpha \beta = \frac { 1 } { 4 } = \frac { c } { a }$
If $a = 4,$ then $b = 1$ and $c = 1.$
So, one quadratic polynomial which fits the given conditions is $4x^2+ x + 1.$
Aliter,
It given that $\alpha + \beta = - \frac { 1 } { 4 } \text { and } \alpha \beta = \frac { 1 } { 4 }$
now, standard form of quadratic polynomial is given by $x ^ { 2 } - ( \alpha + \beta ) x + \alpha \beta$
$= x ^ { 2 } - ( \alpha + \beta ) x + \alpha \beta$
$= x ^ { 2 } - \left( - \frac { 1 } { 4 } \right) x + \frac { 1 } { 4 }$
$= \frac { 1 } { 4 } \left( 4 x ^ { 2 } + x + 1 \right)$
Hence the required quadratic polynomial is $4x^2+ x + 1$

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Question 23 Marks

Find the zeroes of quadratic polynomial $3x^2- x - 4$ and verify the relationship between the zeroes and their coefficients.

Answer

The quadratic equation is given as: $3x^2– x – 4$
$($Now we will factorize $1$ in such a way that the product of factors is equal to $-12$ and the sum is equal to $1)$
$= 3x^2- 4x + 3x - 4$
$= x(3x - 4) + 1(3x - 4)= (3x – 4 )(x + 1)$
The value of $3x^2− x − 4$ is zero when $3x − 4 = 0$ or $x + 1 = 0,$
when $x=\frac{4}{3} \text { or } x=-1$
Therefore, the zeroes of $3x^2− x − 4$ are $\frac{4}{3}$ and $-1$
Sum of zeroes = $\frac{4}{3}+(-1)=\frac{1}{3}=\frac{-(-1)}{3}=\frac{-\text { coefficient of } x}{\text { coefficient of } x^{2}}$
Product of zeroes =$\frac{4}{3}(-1)=\frac{-4}{3}=\frac{\text { constant term }}{\text { coefficient of } x^{2}}$
Hence verified.

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Question 33 Marks

Find the zeroes of quadratic polynomial $x^2- 2x - 8$ and verify the relationship between the zeroes and their coefficients.

Answer

Let $p(x) = x^2- 2x - 8$
By the method of splitting the middle term,
$x ^ { 2 } - 2 x - 8 = x ^ { 2 } - 4 x + 2 x - 8$
$= x ( x - 4 ) + 2 ( x - 4 ) = ( x - 4 ) ( x + 2 )$
For zeroes of $p(x),$
$p(x) = 0 \Rightarrow ( x - 4 ) ( x + 2 ) = 0$
$\Rightarrow x - 4 = 0 \text { or } x + 2 = 0$
$\Rightarrow x = 4 \text { or } x = - 2$
$\Rightarrow x = 4 , - 2$
So, the zeroes of $p(x)$ are $4$ and $-2.$
We observe that, Sum of its zeroes
$= 4 + (-2) = 2$
$= \frac { - ( - 2 ) } { 1 } = \frac { - \text { (Coefficient of } x ) } { \text { Coefficient of } x ^ { 2 } }$
Product of its zeroes
$= 4 x ( - 2 ) = - 8 = \frac { - 8 } { 1 } = \frac { \text { Constant term } } { \text { Coefficient of } x ^ { 2 } }$
Hence, relation between zeroes and coefficients is verified.

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Question 43 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 53 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 63 Marks

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

Answer

Given the zeros are $2+\sqrt{3}$ and $2-\sqrt{3}$, we can find the polynomial by first calculating the sum and product of these zeros.
Let $\alpha=2+\sqrt{3}$ and $\beta=2-\sqrt{3}$.
Sum =$\alpha+\beta=(2+\sqrt{3})+(2-\sqrt{3})=4$
Product =$\alpha \beta=(2+\sqrt{3})(2-\sqrt{3})$
Using the identity $(a+b)(a-b)=a^2-b^2:$
Product =$(2)^2-(\sqrt{3})^2=4-3=1$
The general form of a quadratic polynomial is given by the formula:
$x^2-$ (Sum of Zeros) $x+$ (Product of Zeros)
Substituting the values we calculated:
$x^2-(4) x+(1)$
Therefore, the polynomial is $x^2-4 x+1$.

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Question 73 Marks

Find the zeros of the polynomial x² - 5 and verify the relationship between the zeros and the coefficients.

Answer

To find the zeros of the polynomial $p(x)=x^2-5$, set the polynomial to zero:
$\begin{array}{l}x^2-5=0 \\ x^2=5 \\ x= \pm \sqrt{5}\end{array}$
The zeros are $\sqrt{5}$ and $-\sqrt{5}$
Verification of the Relationship
For a quadratic polynomial of the form $a x^2+b x+c$ the sum of the zeros is given by the formula $-\frac{b}{a}$ and the product of the zeros is given by $\frac{c}{a}$.
Here, the polynomial is $x^2-5$. Comparing this to $a x^2+b x+c$, we have:
$\begin{array}{l}a=1 \\ b=0 \\ c=-5\end{array}$
Let the zeros be $\alpha=\sqrt{5}$ and $\beta=-\sqrt{5}$.
Calculated Sum: $\alpha+\beta=\sqrt{5}+(-\sqrt{5})=0$
$-\frac{b}{a}=-\frac{0}{1}=0$
The relationship is verified as both sums are equal.
Calculated Product: $\alpha \beta=(\sqrt{5})(-\sqrt{5})=-5$
$\frac{c}{a}=\frac{-5}{1}=-5$
The relationship is verified as both products are equal.

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Maths 3 Marks Question

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