Problem of the Week

Updated at Jan 9, 2023 11:10 AM

This week's problem comes from the equation category.

How would you solve the equation \(4(z-3)+{(4z)}^{2}=8\)?

Let's begin!



\[4(z-3)+{(4z)}^{2}=8\]

1
Use Multiplication Distributive Property: \({(xy)}^{a}={x}^{a}{y}^{a}\).
\[4(z-3)+{4}^{2}{z}^{2}=8\]

2
Simplify  \({4}^{2}\)  to  \(16\).
\[4(z-3)+16{z}^{2}=8\]

3
Expand.
\[4z-12+16{z}^{2}=8\]

4
Move all terms to one side.
\[4z-12+16{z}^{2}-8=0\]

5
Simplify  \(4z-12+16{z}^{2}-8\)  to  \(4z-20+16{z}^{2}\).
\[4z-20+16{z}^{2}=0\]

6
Factor out the common term \(4\).
\[4(z-5+4{z}^{2})=0\]

7
Split the second term in \(z-5+4{z}^{2}\) into two terms.
\[4(4{z}^{2}+5z-4z-5)=0\]

8
Factor out common terms in the first two terms, then in the last two terms.
\[4(z(4z+5)-(4z+5))=0\]

9
Factor out the common term \(4z+5\).
\[4(4z+5)(z-1)=0\]

10
Solve for \(z\).
\[z=-\frac{5}{4},1\]

Done

Decimal Form: -1.25, 1
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