Exercise 22 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 and then simplify as much as possible. 81x^2 = 9 ⇔ 81x^2 - 9 = 0 Now, we can identify the values of a, b, and c. 81x^2 - 9 = 0 ⇕ 81x^2+ 0x+( - 9)=0 We see that a= 81, b= 0, and c= - 9. Let's substitute these values into the Quadratic Formula.

x=- b±sqrt(b^2-4ac)/2a

SubstituteValues

Substitute values

x=- 0±sqrt(0^2-4( 81)( - 9))/2( 81)

▼

Solve for x and Simplify

CalcPow

Calculate power

x=- 0±sqrt(0-4(81)(- 9))/2(81)

Multiply

Multiply

x=- 0±sqrt(0-324(- 9))/162

MultNegNegOnePar

- a(- b)=a* b

x=- 0±sqrt(0+2916)/162

AddTerms

Add terms

x = ± sqrt(2916)/162

CalcRoot

Calculate root

x=± 54/162

ReduceFrac

a/b=.a /54./.b /54.

x=± 1/3

The solutions for this equation are x= ± 13. Let's separate them into the positive and negative cases.

x=± 1/3
x_1=1/3	x_2=- 1/3
x_1≈ 0.3	x_2≈ - 0.3

Using the Quadratic Formula, we found that the solutions of the given equation are x_1 ≈ 0.3 and x_2 ≈ - 0.3.