# Math Help - Quadratic Equations

1. ## Quadratic Equations

Hello there, I've got a quadratic equation. I'm supposed to solve it by factorising but it's already factorized.

$(x-2)(2x+1)=0$

Can you help me?

2. ## Re: Quadratic Equations

Use the fact that $a\cdot b=0 \Leftrightarrow a=0 \ \mbox{or} \ b=0$

3. ## Re: Quadratic Equations

For $(3x + 1)(x + 3) = 19 - 2(x +2)$

Do I expand everything on the left hand side then do the same on the the Right hand side then move the right hand side to the left hand side to make it = 0?

4. ## Re: Quadratic Equations

Yep have a go at that.

5. ## Re: Quadratic Equations

After a long complex situation I get

$x^2+10x+18=0$

Which I can't do as nothing multiplys to make 18 thats add to 10

6. ## Re: Quadratic Equations

For $\displaystyle ax^2+bx+c=0\implies x = \frac{-b\pm\sqrt{b^2-4ac}}{2a}$

7. ## Re: Quadratic Equations

But I've got to do via factorisation. Don't think I can factorise the above. Think I done something wrong.

8. ## Re: Quadratic Equations

Complete the square then.

9. ## Re: Quadratic Equations

Originally Posted by JibblyJabbly
But I've got to do via factorisation
$x^2 + 10x + 18 = 0$

if you insist ...

$(x + 5 + \sqrt{7})(x + 5 - \sqrt{7}) = 0$

10. ## Re: Quadratic Equations

Originally Posted by skeeter
$x^2 + 10x + 18 = 0$

if you insist ...

$(x + 5 + \sqrt{7})(x + 5 - \sqrt{7}) = 0$
Just to double check
$(3x + 1)(x + 3) = 19 - 2(x +2)$ is equal to

$x^2 + 10x + 18 = 0$ right? I assume I had problems with the Left hand side of the equation

11. ## Re: Quadratic Equations

Better check my workings but,

$(3x + 1)(x + 3) = 19 - 2(x +2)$

$3x^2+9x+x+3 = 19 - 2x -4$

$3x^2+10x+3 = 15 - 2x$

$3x^2+12x-12 = 0$

$3(x^2+4x-4) = 0$

$x^2+4x-4 = 0$