Question

the length of rectangle exceeds its width by 8cm and area of rectangle is 240sq.cm.find dimension of rectangle. sove this by quadratic equations plz solve it

Answer 1

Area of the rectangle =Length * width

Let x= width of the rectangle

x+8 = Width of the rectangle

Area of the rectangle = 240

Thus x(x+8)=240

x^2+8x =240

Subtracting 240 on both sides,

x^2+8x-240=0

Using Quadratic formula to solve this equation.

$\mathrm{For\:a\:quadratic\:equation\:of\:the\:form\:}ax^2+bx+c=0\mathrm{\:the\:solutions\:are\:}$

$x_{1,\:2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

$\mathrm{For\:}\quad a=1,\:b=8,\:c=-240:\quad x_{1,\:2}=\frac{-8\pm \sqrt{8^2-4\cdot \:1\left(-240\right)}}{2\cdot \:1}$

$x=\frac{-8+\sqrt{8^2+4\cdot \:1\cdot \:240}}{2\cdot \:1}=12$

$x=\frac{-8-\sqrt{8^2+4\cdot \:1\cdot \:240}}{2\cdot \:1}=-20 (impossible)$

So the width of the rectangle =x = 12 cm

Length of the rectangle = x+8=12+8=20 cm