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

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

$\displaystyle x^2+(x+3)^2=11^2$

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".....

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?

$\displaystyle \\x^2+(x+3)^2=11^2\\x^2+x^2+6x+9=121\\2x^2+6x-112=0\\x^2+3x-56=0$

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...