# Quadratic Equations with Pythagoras

• April 2nd 2013, 08:26 AM
janvi anish
Quadratic Equations with Pythagoras
Hi,
I was just wondering if anyone could help me.
I have this maths equation..

A rectangle is 3m longer than it is wide. The diagonal is 11cm. if 'x' is the width of the rectangle then:
Use Pythagoras theorem to find an equation involving 'x' and show that it simplifies to x² + 3x - 56 = 0

Thanks
• April 2nd 2013, 09:15 AM
Plato
Re: Quadratic Equations with Pythagoras
Quote:

Originally Posted by janvi anish
A rectangle is 3m longer than it is wide. The diagonal is 11cm. if 'x' is the width of the rectangle then: Use Pythagoras theorem to find an equation involving 'x' and show that it simplifies to x² + 3x - 56 = 0

$x^2+(x+3)^2=11^2$
• April 2nd 2013, 09:22 AM
janvi anish
Re: Quadratic Equations with Pythagoras
hi thanks..i got this as well but then it does not work>...

x² + 3x - 56 = 0 does not = x" + (x +3)" = 11".....
• April 2nd 2013, 09:40 AM
Plato
Re: Quadratic Equations with Pythagoras
Quote:

Originally Posted by janvi anish
hi thanks..i got this as well but then it does not work>...

x² + 3x - 56 = 0 does not = x" + (x +3)" = 11".....

What do you mean it does not work?
$\\x^2+(x+3)^2=11^2\\x^2+x^2+6x+9=121\\2x^2+6x-112=0\\x^2+3x-56=0$
• April 2nd 2013, 09:41 AM
janvi anish
Re: Quadratic Equations with Pythagoras
oh god...i have been stuck for ages..i added the 9 to the 121 to make it 130...Thanks for spotting the stupid mistake i did..thks...