Exercise 16 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 We first need to identify the values of a, b, and c. 4x^2+5x-6=0 ⇕ 4x^2+ 5x+( - 6)=0 We see that a= 4, b= 5, and c= - 6. Let's substitute these values into the Quadratic Formula.

x=- b±sqrt(b^2-4ac)/2a

SubstituteValues

Substitute values

x=- 5±sqrt(5^2-4( 4)( - 6))/2( 4)

▼

Solve for x and Simplify

CalcPow

Calculate power

x=- 5±sqrt(25-4(4)(- 6))/2(4)

Multiply

Multiply

x=- 5±sqrt(25-16(- 6))/8

MultNegNegOnePar

- a(- b)=a* b

x=- 5±sqrt(25 + 96)/8

AddTerms

Add terms

x=- 5±sqrt(121)/8

CalcRoot

Calculate root

x=- 5± 11/8

The solutions for this equation are x= - 5± 118. Let's separate them into the positive and negative cases.

x=- 5± 11/8
x_1=- 5 + 11/8	x_2=- 5 - 11/8
x_1=6/8	x_2=- 16/8
x_1= 0.75	x_2= - 2

Using the Quadratic Formula, we found that the solutions of the given equation are x_1=0.75 and x_2=- 2.