Quadratic Equations

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September 21st 2011, 01:08 PM
JibblyJabbly

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?
September 21st 2011, 01:15 PM
Siron

Re: Quadratic Equations

Use the fact that $a\cdot b=0 \Leftrightarrow a=0 \ \mbox{or} \ b=0$
September 21st 2011, 03:06 PM
JibblyJabbly

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?
September 21st 2011, 03:23 PM
pickslides

Re: Quadratic Equations

Yep have a go at that.
September 21st 2011, 04:53 PM
JibblyJabbly

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
September 21st 2011, 04:57 PM
pickslides

Re: Quadratic Equations

For $\displaystyle ax^2+bx+c=0\implies x = \frac{-b\pm\sqrt{b^2-4ac}}{2a}$
September 21st 2011, 05:03 PM
JibblyJabbly

Re: Quadratic Equations

But I've got to do via factorisation. Don't think I can factorise the above. Think I done something wrong.
September 21st 2011, 05:09 PM
pickslides

Re: Quadratic Equations

Complete the square then.
September 21st 2011, 05:11 PM
skeeter

Re: Quadratic Equations

Quote:

Originally Posted by JibblyJabbly

But I've got to do via factorisation

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

if you insist ...

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

Re: Quadratic Equations

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
September 21st 2011, 05:46 PM
pickslides

Re: Quadratic Equations

Better check my workings but,

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

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

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

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

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

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

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