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Mathematics · 2022 · 2 marks
CBSE 2022 · Region 4 · Set 1 · Q3
(a) If the sum of the roots of the quadratic equation ky^2-11 y +( k -23)=0 is 13/21 more than the product of the roots, then find the value of k.
(a)
If the sum of the roots of the quadratic equation $\displaystyle \mathrm{ky}^{2}-11 \mathrm{y}+(\mathrm{k}-23)=0$ is $\displaystyle \frac{13}{21}$ more than the product of the roots, then find the value of k.
OR
(b)
If $\displaystyle x=-2$ is the common solution of quadratic equations $\displaystyle \mathrm{a} x^{2}+x-3 \mathrm{a}=0$ and $\displaystyle x^{2}+\mathrm{b} x+\mathrm{b}=0$, then find the value of $\displaystyle \mathrm{a}^{2} \mathrm{~b}$.
Marking-scheme solution
$\displaystyle k y^{2}-11 y+(k-23)=0$. Here $\displaystyle a=k, b=-11, c=k-23$
Sum of roots $\displaystyle =\frac{11}{k}$
Product of roots $\displaystyle =\frac{k-23}{k}$
$\displaystyle \mathrm{ATQ}, \frac{11}{k}=\frac{k-23}{k}+\frac{13}{21}$
Solving, we get $\displaystyle k=21$
Or
If $\displaystyle x=-2$ is the common solution of quadratic equations $\displaystyle \mathrm{a} x^{2}+x-3 \mathrm{a}=0$ and $\displaystyle x^{2}+\mathrm{b} x+\mathrm{b}=0$, then find the value of $\displaystyle \mathrm{a}^{2} \mathrm{~b}$.
$\displaystyle x=-2$ is the common solution of $\displaystyle a x^{2}+x-3 a=0$ and $\displaystyle x^{2}+b x+b=0$.
$\displaystyle \therefore a(-2)^{2}+(-2)-3 a=0 \Rightarrow 4 a-2-3 a=0$
$\displaystyle a=2$
And $\displaystyle (-2)^{2}+b(-2)+b=0 \Rightarrow 4-2 \mathrm{~b}+\mathrm{b}=0 \Rightarrow \mathrm{~b}=4$
$\displaystyle a^{2} b=4 \times 4=16$
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