Find the discriminant of the equation and give the number and type of solutions of the equation?

Question

1) 6x^2=4-5x
2) 2y^2-3y=-4

Answer 1

1) 6x² + 5x - 4 = 0

a = 6
b = 5
c = -4

The discriminant is the portion of the quadratic equation under the sqrt sign. Specifically:

Discriminant = b² - 4ac
= 5² - 4(6)(-4)
= 25 + 96
= 121

Since the discriminant is positive you will have 2 real roots. This is because you can take the sqrt of a positive number and you'll have +/- a real number which means 2 real roots.

2) 2y² - 3y + 4 = 0

a = 2
b = -3
c = 4

Discriminant = (-3)² - 4(2)(4)
= 9 - 32
= -23

Since it is negative (and sqrt of a negative results in an imaginary number) you will have two *imaginary* roots.

(Note: if you ever get zero for the discriminant, that means *one* real root.

Here's the summary:
Discriminant positive --> two real roots
Discriminant negative --> two imaginary roots
Discriminant zero --> one real root

1) discriminant = 121, so two real roots
2) discriminant = -23, so two imaginary roots

Answer 2

First write each equation in the form ax^2 + bx + c = 0
1) 6x^2 = 4 - 5x
Add 5x to each side and subtract 4 from each side.
6x^2 + 5x - 4 = 0
So a = 6, b = 5 and c = -4
The discriminant is b^2 - 4ac
= (5)^2 - 4(6)(-4)
= 25 + 96
= 121 >0
Therefore, there are two distinct roots.
Note: If the discriminant is zero, then there is one real root.
If the discriminant is negative, then the roots are complex.
2) 2y^2 - 3y = -4
Add 4 to both sides
2y^2 - 3y + 4 = 0
So a = 2, b = -3 and c = 4
b^2 - 4ac = (-3)^2 - 4(2)(4)
= 9 - 32
= -23 < 0
Therefore, the roots are complex (or you could say they are not real)

Answer 3

no