Review question

# Can we find the repeated root for $4x^4+x^2+3x+1=0$? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource

Ref: R9082

## Solution

Prove that, if $\alpha$ is a repeated root of $f(x)=0$, where $f(x)$ is a polynomial, then $\alpha$ is a root of the equation $f'(x)=0$.

Since we know that $f(x)=0$ has a repeated root $\alpha$, we may write $f(x)=(x-\alpha)^2 g(x),$ where $g(x)$ is some polynomial. By the product rule we have $f'(x)=2(x-\alpha)g(x)+(x-\alpha)^2 g'(x)=(x-\alpha)\left[2g(x)+(x-\alpha)g'(x)\right],$ so we have that $\alpha$ is a root of $f'(x)=0$.

Given that the equation $4x^4+x^2+3x+1=0$ has a repeated root, find its value.

If $f(x)=4x^4+x^2+3x+1$, then $f'(x)=16x^3+2x+3.$

We are told that $f(x) = 0$ has a repeated root, which we may call $\alpha$. From the above, we now know that $\alpha$ is a root of $f'(x) = 0$.

So we seek a root of $16x^3+2x+3=0$.

Clearly the root $x$ can’t be a whole number, since $16x^3+2x$ would be even and $3$ odd.

The coefficients $16$ and $2$ make us think the substitution $x = \dfrac{y}{2}$ would be helpful, where $y$ is a whole number.

That gives $2y^3 + y + 3 = 0$, and clearly $y = -1$ is a root here. We have $2y^3 + y + 3 = (y + 1)(2y^2 -2y + 3)$.

Note the roots of the quadratic bracket are complex, since $b^2 < 4ac$.

So the repeated root we seek must be $-\dfrac{1}{2}$. We can check this, finding

$f(x) = (2x + 1)(2x^3 - x^2 + x + 1) = (2x + 1)^2(x^2 - x + 1).$

Alternatively, we could note that $f'(0)$ = 3, and $f'(-1) = -15$, so $f'(x) = 0$ must have a root between $-1$ and $0$, since $f'(x)$ is continuous and we have a change of sign.

The value $-1/2$ is a sensible choice to test as a possible root, and it works.