• Sep 21st 2011, 01:08 PM
JibblyJabbly
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?
• Sep 21st 2011, 01:15 PM
Siron
Use the fact that $a\cdot b=0 \Leftrightarrow a=0 \ \mbox{or} \ b=0$
• Sep 21st 2011, 03:06 PM
JibblyJabbly
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?
• Sep 21st 2011, 03:23 PM
pickslides
Yep have a go at that.
• Sep 21st 2011, 04:53 PM
JibblyJabbly
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
• Sep 21st 2011, 04:57 PM
pickslides
For $\displaystyle ax^2+bx+c=0\implies x = \frac{-b\pm\sqrt{b^2-4ac}}{2a}$
• Sep 21st 2011, 05:03 PM
JibblyJabbly
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 PM
pickslides
Complete the square then.
• Sep 21st 2011, 05:11 PM
skeeter
Quote:

Originally Posted by JibblyJabbly
But I've got to do via factorisation

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

if you insist ...

$(x + 5 + \sqrt{7})(x + 5 - \sqrt{7}) = 0$
• Sep 21st 2011, 05:26 PM
JibblyJabbly
Quote:

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
• Sep 21st 2011, 05:46 PM
pickslides
Better check my workings but,

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

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

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

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

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

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