Exercise 37 Page 137

We want to use the discriminant of the given quadratic equation to determine the number of real solutions. In the Quadratic Formula, b^2-4ac is the discriminant. ax^2+bx+c=0 ⇕ x=- b±sqrt(b^2-4ac)/2a If we just want to know the number of real solutions, and not the solutions themselves, we only need to work with the discriminant. Let's first rewrite the given equation in standard form. x^2 - 4/5x = 3 ⇔ x^2 - 4/5x - 3 = 0 Having rewritten the equation, we can now identify the values of a, b, and c. x^2 - 4/5x - 3 = 0 ⇕ 1x^2+( - 4/5)x+( - 3)=0 Finally, let's evaluate the discriminant.

b^2-4ac

SubstituteValues

Substitute values

( - 4/5)^2-4( 1)( -3)

▼

Simplify

NegBaseToPosBase

(- a)^2=a^2

(4/5)^2 - 4(1)(- 3)

ExpandFrac

a/b=a * 2/b * 2

(8/10)^2 - 4(1)(- 3)

Rewrite

Rewrite 8/10 as 0.8

(0.8)^2 - 4(1)(- 3)

CalcPow

Calculate power

0.64 - 4(1)(-3)

MultByOne

a * 1=a

0.64 - 4(- 3)

MultNegNegOnePar

- a(- b)=a* b

0.64 + 12

AddTerms

Add terms

12.64

Since the discriminant is 12.64, the quadratic equation has two real solutions.

Extra

Further information

If the discriminant is greater than zero, the equation will have two real solutions. If it is equal to zero, the equation will have one real solution. Finally, if the discriminant is less than zero, the equation will have no real solutions.