2

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.

Xander Henderson
  • 29,772
brilliant
  • 818

5 Answers5

4

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 $b=-3$.

Xander Henderson
  • 29,772
MPW
  • 43,638
  • 1
    This is just amazing! Thank you. So simple and yet i myself would have never guessed. it looks like math is more than just memorizing and counting. It is also about intuition in finding solution, sometimes non-standard ones. – brilliant Dec 30 '19 at 12:57
  • Glad to help. The thought process here is often “How can I get from what I’m given to what I need?” I see a squared $x$ in the target but not in the source, so I suspect I need to square something. And I also see that nasty root thingy which will go away if I square, so I am feeling pretty strongly that I need to square things. I also then imagine that after I do that, I will see $x^2$ but not $5x^2$, so that makes me want to multiply by $5$ to make the match even closer. Then just massage it a little to get what you need. You’ll learn this with practice. – MPW Dec 30 '19 at 14:13
  • Very useful. Thank you! – brilliant Dec 30 '19 at 14:50
3

$(x-\frac 4 5)^{2}=\frac {31} {25}$ 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 $b =-3$.

Kavi Rama Murthy
  • 311,013
1

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 $$x^2-\frac85 x-\frac35=0\iff 5x^2-8x-3=0.$$

Bernard
  • 175,478
1

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

1

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: $b=-3$

Matteo
  • 6,581