Exercise 10 Page 527

We want to solve the quadratic equation by completing the square. To do so, we will start by rewriting the equation so only terms with p are on one side of the equation. p^2+10p -4 =0 ⇔ p^2+10p=4In a quadratic expression, b is the linear coefficient. For the equation above, we have that b=10. Let's now calculate ( b2 )^2.

( b/2 )^2

Substitute

b= 10

( 10/2 )^2

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Simplify

CalcQuot

Calculate quotient

5^2

CalcPow

Calculate power

25

Next, we will add ( b2 )^2=25 to both sides of our equation. Then, we will factor the trinomial on the left-hand side and solve the equation.

p^2+10p=4

AddEqn

LHS+25=RHS+25

p^2+10p + 25 = 4 + 25

FacPosPerfectSquare

a^2+2ab+b^2=(a+b)^2

(p+5)^2=4+25

AddTerms

Add terms

(p+5)^2=29

SqrtEqn

sqrt(LHS)=sqrt(RHS)

sqrt((p+5)^2)=sqrt(29)

CalcRoot

Calculate root

p + 5 =± 5.385164 ⋯

RoundDec

Round to 2 decimal place(s)

p + 5 ≈ ± 5.39

SubEqn

LHS-5=RHS-5

p ≈ -5 ± 5.39

Both p ≈ - 10.39 and p ≈ 0.39 are solutions of the equation.