Question

1. Evaluate the discriminant for each equation, then use it to determine the nature of the roots for each quadratic equation.

(a) x^+4x+5=0

(b) x^-4x-5=0

(c) 4x^2+20x+25=0

Answer 1

a. x^2+4x+5.
We first have to identify the values for a, b, c.

In this equation:
a=1
b=4
c=4

Then we have to plug the values into the discriminant equation.
The Equation for the discriminant is Sqrt b^2-4ac

A good rule to remember is If the discriminant is greater than zero, there are two solutions. If the discriminant is less than zero, there are no solutions and if the discriminant is equal to zero, there is one solution.

4^2-4(1)(4) = 0

a. Discriminant is 0 and there is one solution.

b. x^2-4x-5=0

a=1
b=-4
c=-5

Sqrt -4^2-4(1)(-5)= 36

The discriminant is greater than zero so there are 2 solutions.

b. Discriminant is 36 and there are 2 solutions.

c. 4x^2+20x+25=0

a=4
b=20
c=25

sqrt 20^2-4(4)(25)=0

The discriminant is 0 so there is only 1 solution.

c. Discriminant is 0 and there is 1 solution

Answer 2

Given the quadratic equation ax^2 + bx + c = 0, the coefficients of x^2, x and 1 are a, b and c respectively.

(a) and (b) are not valid quadratics, because they omit the exponent 2. Rewrite these as comments within your original post, please.

(c) 4x^2+20x+25=0: Here, a = 4, b = 20 and c = 25.