Exercise 5 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. 10x^2-31x + 15 = 0 ⇕ 10x^2+( - 31)x+ 15=0 We see that a= 10, b= - 31, and c= 15. Let's substitute these values into the Quadratic Formula.

x=- b±sqrt(b^2-4ac)/2a

SubstituteValues

Substitute values

x=- ( -31)±sqrt(( - 31)^2-4( 10)( 15))/2( 10)

▼

Solve for x and Simplify

NegNeg

- (- a)=a

x=31±sqrt((- 31)^2-4(10)(15))/2(10)

CalcPow

Calculate power

x=31±sqrt(961-4(10)(15))/2(10)

Multiply

Multiply

x=31±sqrt(961-600)/20

SubTerm

Subtract term

x=31±sqrt(361)/20

CalcRoot

Calculate root

x=31 ± 19/20

The solutions for this equation are x= 31 ± 1920. Let's separate them into the positive and negative cases.

x=31± 19/20
x_1=31-19/20	x_2=31 + 19/20
x_1=12/20	x_2=50/20
x_1=0.6	x_2=2.5

Using the Quadratic Formula, we found that the solutions of the given equation are x_1=0.6 and x_2=2.5.