Review question

# Can we find both roots if one is double the other? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource

Ref: R6638

## Solution

Given that one of the values of $x$ satisfying the quadratic equation $x^2-(2a+1)x+(a^2+2)=0$ is twice the other, find $a$.

If $x^2-(2a+1)x+a^2+2 = 0$ has one root double the other, then we can call the roots $b$ and $2b$, and write $x^2-(2a+1)x+a^2+2 \equiv (x-b)(x-2b),$ as the quadratic on the right hand side has roots $b$ and $2b$.

Multiplying out we have $x^2-(2a+1)x+a^2+2 \equiv x^2-3bx+2b^2.$

Since this holds for all $x$ (it is an identity), we can equate coefficients, giving us $2a+1=3b$ and $a^2 + 2 =2b^2$.

The first equation gives $b = \dfrac{2a+1}{3}$, so we can substitute this into the second equation to obtain $a^2 + 2 = 2\left(\dfrac{2a+1}{3}\right)^2.$

Expanding the brackets and rearranging yields $a^2 - 8a + 16 = 0$, or $(a-4)^2=0$, so $a = 4$.

Checking this in the original equation: $x^2-(2 \times 4+1)x+4^2+2 = (x-3)(x-6)$.