Question

Given $\alpha$ and $\beta$ are the real roots of the quadratic equation $x^2-4x+k=0(k\ne 0)$. If ${\alpha \beta }^2+{\alpha }^2\beta$.

• A. $4k$
• B. $\frac{16}{7}$
• C. $\frac{3}{7}$
• D. $12$

Answer 1

Answer: (A)

$4k$

$x^2-4x+k=0$

$\Rightarrow$ Here, $a=1,b=-4,c=k$

Let roots of the equation are real and equal.

$\therefore$ $b^2-4ac=0$

$\Rightarrow$ $(-4{)}^2-4(1\left)\right(k)=0$

$\Rightarrow$ $16-4k=0$

$\Rightarrow$ $4k=16$

$\therefore$ $k=4$

Substituting value of $k$ in given equation we get,

$\Rightarrow$ $x^2-4x+4=0$

$\Rightarrow$ $x^2-2x-2x+4=0$

$\Rightarrow$ $x(x-2)-2(x-2)=0$

$\Rightarrow$ $(x-2)(x-2)=0$

$\therefore$ $x=2,2$

$\therefore$ $\alpha =2$ and $\beta =2$

$\Rightarrow$ $\alpha {\beta }^2+{\alpha }^2\beta =\left(2\right)(2{)}^2+(2{)}^2\left(2\right)$

$=2\times 4+4\times 2$

$=4(2+2)$

$=4\left(k\right)$ [ Since $k=4$ ]

$\therefore$ $\alpha {\beta }^2+{\alpha }^2\beta =4k$