Exercise 11 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 k are on one side of the equation. k^2-4k-7 =0 ⇔ k^2-4k=7In a quadratic expression, b is the linear coefficient. For the equation above, we have that b=-4. Let's now calculate ( b2 )^2.

( b/2 )^2

Substitute

b= -4

( -4/2 )^2

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Simplify

CalcQuot

Calculate quotient

(-2)^2

CalcPow

Calculate power

4

Next, we will add ( b2 )^2=4 to both sides of our equation. Then, we will factor the trinomial on the left-hand side and solve the equation.

k^2-4k=7

AddEqn

LHS+4=RHS+4

k^2-4k + 4 = 7 + 4

FacNegPerfectSquare

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

(k-2)^2=7+4

AddTerms

Add terms

(k-2)^2=11

SqrtEqn

sqrt(LHS)=sqrt(RHS)

sqrt((k-2)^2)=sqrt(11)

CalcRoot

Calculate root

k-2 =± 3.316624 ⋯

RoundDec

Round to 2 decimal place(s)

k-2 ≈ ± 3.32

SubEqn

LHS-5=RHS-5

k ≈ 2 ± 3.32

Both k ≈ - 1.32 and k ≈ 5.32 are solutions of the equation.