# 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 $$\frac{{({p}^{2}-3)}^{2}}{5}=\frac{1}{5}$$?

Let's begin!

$\frac{{({p}^{2}-3)}^{2}}{5}=\frac{1}{5}$

 1 Multiply both sides by $$5$$.${({p}^{2}-3)}^{2}=\frac{1}{5}\times 5$2 Cancel $$5$$.${({p}^{2}-3)}^{2}=1$3 Take the square root of both sides.${p}^{2}-3=\pm \sqrt{1}$4 Simplify  $$\sqrt{1}$$  to  $$1$$.${p}^{2}-3=\pm 1$5 Break down the problem into these 2 equations.${p}^{2}-3=1$${p}^{2}-3=-1$6 Solve the 1st equation: $${p}^{2}-3=1$$.1 Add $$3$$ to both sides.${p}^{2}=1+3$2 Simplify  $$1+3$$  to  $$4$$.${p}^{2}=4$3 Take the square root of both sides.$p=\pm \sqrt{4}$4 Since $$2\times 2=4$$, the square root of $$4$$ is $$2$$.$p=\pm 2$To get access to all 'How?' and 'Why?' steps, join Cymath Plus!$p=\pm 2$7 Solve the 2nd equation: $${p}^{2}-3=-1$$.1 Add $$3$$ to both sides.${p}^{2}=-1+3$2 Simplify  $$-1+3$$  to  $$2$$.${p}^{2}=2$3 Take the square root of both sides.$p=\pm \sqrt{2}$To get access to all 'How?' and 'Why?' steps, join Cymath Plus!$p=\pm \sqrt{2}$8 Collect all solutions.$p=\pm 2,\pm \sqrt{2}$DoneDecimal Form: ±2, ±1.414214p=2,-2,sqrt(2),-sqrt(2)