Question

A rectangle with an area of 24 square units has length x+1 and width 4x-6. find the value of x.

Answer 1

$area = length \cdot width \\ \\A = 24 s^2 \\ \\ l=x+1 , \ \ w=4x-6 \\ \\ A=lw\\ \\24 =(x+1)(4x-6)$

$24 =4x^2-6x+4x-6 \\ \\4x^2-6x+4x-6 -24=0\\ \\4x^2-2x-30=0 \ \ / :2\\ \\2x^2-x-15=0 \\ \\a=2, \ b=-1 , c= -15$
 
$\Delta =b^2-4ac = (-1)^2 -4\cdot2\cdot (-15) = 1+120=121\\ \\x_{1}=\frac{-b-\sqrt{\Delta} }{2a}=\frac{1-\sqrt{121}}{2*2 }=\frac{ 1-11}{4}=\frac{-10}{4}=- \frac{5}{2} \\ \\x_{2}=\frac{-b+\sqrt{\Delta} }{2a}=\frac{1+\sqrt{121}}{2*2 }=\frac{ 1+11}{4}=\frac{12}{4}=3 \\ \\Use \ positive \ value \ of \ x : \\ \\ x =3$