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

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

Can you help me?

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- Sep 21st 2011, 01:08 PMJibblyJabblyQuadratic Equations
Hello there, I've got a quadratic equation. I'm supposed to solve it by factorising but it's already factorized.

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

Can you help me? - Sep 21st 2011, 01:15 PMSironRe: Quadratic Equations
Use the fact that $\displaystyle a\cdot b=0 \Leftrightarrow a=0 \ \mbox{or} \ b=0$

- Sep 21st 2011, 03:06 PMJibblyJabblyRe: Quadratic Equations
For $\displaystyle (3x + 1)(x + 3) = 19 - 2(x +2)$

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? - Sep 21st 2011, 03:23 PMpickslidesRe: Quadratic Equations
Yep have a go at that.

- Sep 21st 2011, 04:53 PMJibblyJabblyRe: Quadratic Equations
After a long complex situation I get

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

Which I can't do as nothing multiplys to make 18 thats add to 10 - Sep 21st 2011, 04:57 PMpickslidesRe: Quadratic Equations
For $\displaystyle \displaystyle ax^2+bx+c=0\implies x = \frac{-b\pm\sqrt{b^2-4ac}}{2a}$

- Sep 21st 2011, 05:03 PMJibblyJabblyRe: Quadratic Equations
But I've got to do via factorisation. Don't think I can factorise the above. Think I done something wrong.

- Sep 21st 2011, 05:09 PMpickslidesRe: Quadratic Equations
Complete the square then.

- Sep 21st 2011, 05:11 PMskeeterRe: Quadratic Equations
- Sep 21st 2011, 05:26 PMJibblyJabblyRe: Quadratic Equations
- Sep 21st 2011, 05:46 PMpickslidesRe: Quadratic Equations
Better check my workings but,

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

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

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

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

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

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