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 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 33 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 43 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 53 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 63 Marks

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

Answer

Let $p(x)=x^2-2 x-8$
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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