Question

the area of a rectangle is 20 cm². the length is 8 more than the width. use w for width. what is the width in cm? equation: w² + 8w - 20 = 0 solution: ? enter the solution. like t=3

Explanation:

Step1: Identify the quadratic equation

We have the quadratic equation \( w^{2}+8w - 20=0 \). To solve this quadratic equation \( ax^{2}+bx + c = 0\) (here \(a = 1\), \(b=8\), \(c=- 20\)), we can use the quadratic formula \(w=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\) or factor the equation. Let's try factoring first. We need two numbers that multiply to \(a\times c=1\times(- 20)=-20\) and add up to \(b = 8\). The numbers are \(10\) and \(- 2\) since \(10\times(-2)=-20\) and \(10+(-2) = 8\).

Step2: Factor the quadratic equation

Rewrite the middle term using the two numbers we found:
\(w^{2}+10w-2w - 20=0\)
Group the terms:
\((w^{2}+10w)-(2w + 20)=0\)
Factor out the common factors from each group:
\(w(w + 10)-2(w + 10)=0\)
Now, factor out \((w + 10)\):
\((w - 2)(w + 10)=0\)

Step3: Solve for \(w\)

Set each factor equal to zero:
\(w-2=0\) or \(w + 10=0\)
For \(w-2=0\), we get \(w = 2\)
For \(w + 10=0\), we get \(w=-10\)

Since the width of a rectangle cannot be negative, we discard \(w=-10\). So the width \(w = 2\).

Answer:

\(w = 2\)