Question

If sum of the two roots of the quadratic equation x2 - ax + b =0 is same as the twice
of their difference, then
(a)3a2 = b
(b)a2 = 16b
(c)3a2 = 26
(d)3a2 = 16b​

Answer 1

Let the roots of the quadratic equation $\sf{x^2-ax+b=0}$ be $\alpha$ and $\beta$ where $\alpha\ \textgreater\ \beta.$ Then we have,

• $\sf{\alpha+\beta=a}$

• $\sf{\alpha\beta=b}$

Because for a quadratic equation $\sf{px^2+qx+r=0\quad (p\neq0)}$ having roots $\alpha\ \&\ \beta,$

• $\sf{\alpha+\beta=-\dfrac{q}{p}}$

• $\sf{\alpha\beta=\dfrac{r}{p}}$

Then, given,

$\longrightarrow\sf{\alpha+\beta=2(\alpha-\beta)}$

$\longrightarrow\sf{a=2\sqrt{(\alpha-\beta)^2}}$

$\longrightarrow\sf{a=2\sqrt{\alpha^2+\beta^2-2\alpha\beta}}$

$\longrightarrow\sf{a=2\sqrt{\alpha^2+\beta^2+2\alpha\beta-4\alpha\beta}}$

$\longrightarrow\sf{a=2\sqrt{(\alpha+\beta)^2-4\alpha\beta}}$

$\longrightarrow\sf{a=2\sqrt{a^2-4b}}$

$\longrightarrow\sf{a^2=4(a^2-4b)}$

$\longrightarrow\sf{a^2=4a^2-16b}$

$\longrightarrow\sf{\underline{\underline{3a^2=16b}}}$