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the area of a rectangle is $20\\ \\mathrm{cm}^2$. the length is 8 more …

Question

the area of a rectangle is $20\\ \mathrm{cm}^2$. the length is 8 more than the width. use $w$ for width. what is the width in cm? equation: $w^2 + 8w - 20 = 0$ solution:?

Explanation:

Step1: Identify the quadratic formula

The quadratic equation is \(w^{2}+8w - 20=0\), where \(a = 1\), \(b = 8\), \(c=- 20\). The quadratic formula is \(w=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\).

Step2: Calculate the discriminant

First, calculate the discriminant \(D=b^{2}-4ac=(8)^{2}-4\times1\times(-20)=64 + 80 = 144\).

Step3: Apply the quadratic formula

Substitute \(a = 1\), \(b = 8\), \(D = 144\) into the quadratic formula:
\(w=\frac{-8\pm\sqrt{144}}{2\times1}=\frac{-8\pm12}{2}\)
We have two solutions:
\(w_1=\frac{-8 + 12}{2}=\frac{4}{2}=2\)
\(w_2=\frac{-8-12}{2}=\frac{-20}{2}=-10\)
Since width cannot be negative, we discard \(w_2\).

Answer:

The width is \(2\) cm.