Question

the length of the rectangle is 5 cm less than twice its width. the area of the rectangle is 42 cm². what is the width of the rectangle? enter the answer

Explanation:

Step1: Define variables

Let the width of the rectangle be $w$ cm. Then the length $l=(2w - 5)$ cm.

Step2: Set up area formula

The area of a rectangle $A = l\times w$. Substituting the values of $l$ and $A$, we get $42=(2w - 5)\times w$.
Expanding gives $42 = 2w^{2}-5w$.
Rearranging to standard quadratic - form: $2w^{2}-5w - 42=0$.

Step3: Solve the quadratic equation

For a quadratic equation $ax^{2}+bx + c = 0$ (here $a = 2$, $b=-5$, $c = - 42$), we can use the quadratic formula $x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$ or factor.
Factor $2w^{2}-5w - 42$:
$2w^{2}-5w - 42=2w^{2}-12w + 7w-42=2w(w - 6)+7(w - 6)=(2w + 7)(w - 6)=0$.
Setting each factor equal to zero:
$2w+7 = 0$ gives $w=-\frac{7}{2}$; $w - 6=0$ gives $w = 6$.
Since the width cannot be negative, we discard $w=-\frac{7}{2}$.

Answer:

$6$ cm