How to solve this quadratic equation (with $x$ represented by a fraction containing a square root)?

Question

If $x-\frac{4}{5} = \pm \frac{\sqrt{31}}{5}$, how can I find the values of a and b in the following equation?

$$5x^2+ax+b=0?$$

I've tried substituting $(\frac{4}{5} \pm \frac{\sqrt{31}}{5})$ for $x$ and got totally confused there.

Answer 1

Square both sides, multiply both sides by $5$, then move all terms to the left side. Note that your original equation is really two equations, but after squaring they are the same:

\begin{align}x-\tfrac45=\pm\tfrac{\sqrt{31}}{5} &\implies \left(x-\tfrac45\right)^2=\left(\pm\tfrac{\sqrt{31}}{5}\right)^2 \\ &\implies x^2-\tfrac85 x + \tfrac{16}{25}=\tfrac{31}{25} \\ &\implies 5x^2-8 x + \tfrac{16}{5}=\tfrac{31}{5} \\ &\implies 5x^2-8 x - \tfrac{15}{5} = 0 \\ &\implies 5x^2-8 x - 3 = 0. \end{align}

So $a=-8$ and $b=-3$.

Answer 2

$(x-\frac 4 5)^{2}=\frac {31} {25}$ can be written as $5x^{2}-8x-3=0$. For the quadratic equation to have exactly these two roots we must have $a =-8$ and $b =-3$.

Answer 3

If $s$ and $p$ are the sum and product of the roots of a quadratic equation, the equation can be written as $$x^2-sx+p=0 $$ up to a non-zero factor, from a well-known result from high school.

Here we have
$$s=2\cdot\frac45=\frac85,\qquad p=\biggl(\frac45-\frac{\sqrt{31}}5\biggr)\biggl(\frac45+\frac{\sqrt{31}}5\biggr)=-\frac{15}{ 25}=-\frac35,$$ whence the equation $$x^2-\frac85 x-\frac35=0\iff 5x^2-8x-3=0.$$

Answer 4

Since the roots are $4/5\pm\sqrt{31}/5$, we have that $x^2+(a/5)x+b/5=(x-(4/5+\sqrt{31}/5))(x-(4/5-\sqrt{31}/5))$.

So, $x^2+(a/5)x+b/5=x^2-(8/5)x+(16/25-31/25)=x^2-(8/5)x-3/5$.

Hence $a=-8, b=-3$.

Answer 5

So, I have: $$x=\frac{-a\pm\sqrt{b^2-20a}}{10}=-\frac{a}{10}\pm\frac{\sqrt{b^2-20a}}{5}$$ Now $x=\frac{4}{5}\pm\frac{\sqrt{31}}{5}$, so: $-\frac{a}{10}=\frac{4}{5}$ and $\frac{\sqrt{b^2-20a}}{5}=\frac{\sqrt{31}}{5}$

From the first $a=-8$ and substituing in the second: $b=-3$