Can we find both roots if one is double the other?

Can we find both roots if one is double the other?

Question

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 $a$.

Suggestion

If we know the values of $x$ that satisfy a quadratic equation, how can we use this to write the equation down?

Could we now compare our equation with the one given?

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Download GeoGebra file

The green curve here is $y=x^2-(2a+1)x+(a^2+2)$.

The blue line segment $OP$ represents the smaller root, and the length $OP'$ is double $OP$.

Solution

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: $x^2-(2 \times 4+1)x+4^2+2 = (x-3)(x-6)$.