Exercise 3 Page 137

We will use the Quadratic Formula to solve the given quadratic equation. ax^2+ bx+ c=0 ⇕ x=- b± sqrt(b^2-4 a c)/2 a Let's start by rewriting the equation so all of the terms are on the left-hand side. x^2 - 8x = - 10 ⇔ x^2 - 8x + 10 = 0 Now, we can identify the values of a, b, and c. x^2 - 8x + 10 = 0 ⇕ 1x^2+( - 8)x+ 10=0 We see that a= 1, b= - 8, and c= 10. Let's substitute these values into the Quadratic Formula.

x=- b±sqrt(b^2-4ac)/2a

SubstituteValues

Substitute values

x=- ( -8)±sqrt(( - 8)^2-4( 1)( 10))/2( 1)

▼

Solve for x and Simplify

NegNeg

- (- a)=a

x=8±sqrt((- 8)^2-4(1)(10))/2(1)

CalcPow

Calculate power

x=8±sqrt(64-4(1)(10))/2(1)

MultByOne

a * 1=a

x=8±sqrt(64-4(10))/2

Multiply

Multiply

x=8±sqrt(64-40)/2

SubTerm

Subtract term

x=8±sqrt(24)/2

SplitIntoFactors

Split into factors

x=8±sqrt(4* 6)/2

SqrtProd

sqrt(a* b)=sqrt(a)*sqrt(b)

x=8± sqrt(4)* sqrt(6)/2

CalcRoot

Calculate root

x=8 ± 2 sqrt(6)/2

FactorOut

Factor out 2

x=2(4 ± sqrt(6))/2

CancelCommonFac

Cancel out common factors

x=4± sqrt(6)

The solutions for this equation are x=4 ± sqrt(6). Let's separate them into the positive and negative cases.

x=4 ± sqrt(6)
x_1=4 + sqrt(6)	x_2=4 - sqrt(6)
x_1 ≈ 4 + 2.4	x_2 ≈ 4 - 2.4
x_1 ≈ 6.4	x_2 ≈ 1.6

Using the Quadratic Formula, we found that the solutions of the given equation are x_1≈ 6.4 and x_2≈ 1.6.