Question

Find the value of a for which one root of the equation x^2+(2a-1)x+a^2+2=0 is twice as large as the other

Answer 1

GIVEN :

The equation $x^2+(2a-1)x+a^2+2=0$ and its root is twice as large as the other

TO FIND :

The value of a in the given equation

SOLUTION :

Given equation $x^2+(2a-1)x+a^2+2=0$ and its root is twice as large as the other

Let $\alpha$ and $2\alpha$ be the two roots (by given)

$Sum of the roots=\alpha+2\alpha=-\frac{b}{a}$

$3\alpha=-\frac{(2a-1)}{1}$

$3\alpha=-(2a-1)$

Squaring on both the sides

$(3\alpha)^2=(-(2a-1))^2$

$3^2\alpha^2=(2a-1)^2$

$9\alpha^2=4a^2-4a+1\hfill (1)$

$Product of the roots=\alpha\times 2\alpha=\frac{c}{a}$

$2\alpha^2=\frac{a^2+2}{1}$

$2\alpha^2=a^2+a\hfill (2)$

Dividing the equations (1) by (2) we get

$\frac{9\alpha^2}{2\alpha^2}=\frac{4a^2-4a+1}{a^2+2}$

$\frac{9}{2}=\frac{4a^2-4a+1}{a^2+2}$

$9\times (a^2+2)=(4a^2-4a+1)\times 2$

$9a^2+18=8a^2-8a+2$

$9a^2+18-8a^2+8a-2=0$

$a^2+8a+16=0$

$(a+4)^2=0$

a+4=0

⇒ a=-4,-4 are roots

Now we can substitute the value of a in the given equation we get

$x^2+(2(-4)-1)x+(-4)^2+2=0$

$x^2+(-8-1)x+4^2+2=0$

$x^2-9x+16+2=0$

$x^2-9x+18=0$

(x-3)(x-6)=0

x-3=0 or x-6=0

∴ x=3 and x=6 are the roots.

Since $\alpha$ and $2\alpha$ be the two roots (by given)

∴ $\alpha=3$ and $2\alpha=2(3)=6$ are the two roots.

∴ the value of a is -4.