(a)OR(b)

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.

$\displaystyle k y^{2}-11 y+(k-23)=0$. 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 $\displaystyle (-2)^{2}+b(-2)+b=0 \Rightarrow 4-2 \mathrm{~b}+\mathrm{b}=0 \Rightarrow \mathrm{~b}=4$ $\displaystyle a^{2} b=4 \times 4=16$