Question

If the roots of x ^ 2 + ax + b = 0 are of the form a.k+a and the roots of x ^ 2 + bx + a = 0 are of the form beta, k + beta where k in R^ + , a and baredifferent then a + b + 8 is equal to​

Answer 1

For the quadratic equation having $x=\alpha,\alpha+k$ as roots, we know that,

$\hookrightarrow2\alpha+k=-a,\ \alpha(\alpha+k)=b,\ a\neq b,\ k\in\mathbb{R}^{+}$

For the quadratic equation having $x=\beta,\beta+k$ as roots, we know that,

$\hookrightarrow2\beta+k=-b,\ \beta(\beta+k)=a,\ a\neq b,\ k\in\mathbb{R}^{+}$

We can find that,

$\hookrightarrow2\alpha+k+\beta(\beta+k)=0,\ 2\beta+k+\alpha(\alpha+k)=0$

So,

$\hookrightarrow2\alpha+k+\beta(\beta+k)=2\beta+k+\alpha(\alpha+k)$

$\hookrightarrow2(\alpha-\beta)+\beta^{2}-\alpha^{2}+k(\beta-\alpha)=0$

$\hookrightarrow2(\alpha-\beta)-(\alpha+\beta)(\alpha-\beta)-k(\alpha-\beta)=0$

$\hookrightarrow(\alpha-\beta)(2-\alpha-\beta-k)=0$

As we know $\alpha-\beta\neq0$, we know that,

$\hookrightarrow \alpha+\beta=-k+2$

So, the required answer is,

$\hookrightarrow\red{\boxed{\red{\alpha+\beta+8=\red{\underline{-k+10}}}}}$