Problem of the Week

Updated at Feb 14, 2022 8:30 AM

This week's problem comes from the equation category.

How can we solve the equation (p2−3)25=15\frac{{({p}^{2}-3)}^{2}}{5}=\frac{1}{5}?

Let's begin!



(p2−3)25=15\frac{{({p}^{2}-3)}^{2}}{5}=\frac{1}{5}

1
Multiply both sides by 55.
(p2−3)2=15×5{({p}^{2}-3)}^{2}=\frac{1}{5}\times 5

2
Cancel 55.
(p2−3)2=1{({p}^{2}-3)}^{2}=1

3
Take the square root of both sides.
p2−3=±1{p}^{2}-3=\pm \sqrt{1}

4
Simplify  1\sqrt{1}  to  11.
p2−3=±1{p}^{2}-3=\pm 1

5
Break down the problem into these 2 equations.
p2−3=1{p}^{2}-3=1
p2−3=−1{p}^{2}-3=-1

6
Solve the 1st equation: p2−3=1{p}^{2}-3=1.
p=±2p=\pm 2

7
Solve the 2nd equation: p2−3=−1{p}^{2}-3=-1.
p=±2p=\pm \sqrt{2}

8
Collect all solutions.
p=±2,±2p=\pm 2,\pm \sqrt{2}

Done

Decimal Form: ±2, ±1.414214
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