Question

Suppose you are solving two different quadratic equations using the quadratic

formula. The first has the following solutions:
-12 ± √144 - 120
6
The second has the following solutions:
-10√100 - 192
8
Explain whether each equation will have real or non-real solutions and why.

Answer 1

Explanation:

To determine whether each equation will have real or non-real solutions, we need to analyze the discriminant of the quadratic formula.

The quadratic formula for a quadratic equation of the form

ax 2 +bx+c=0 is:

x= −b± √b2 −4ac ÷ 2a

​

The discriminant (D) is the value inside the square root, i.e., b2 −4ac. The nature of the solutions depends on the value of the discriminant:

1. D>0, the quadratic equation has two distinct real solutions.

2. D=0, the quadratic equation has one real solution (a repeated root).

3. D<0, the quadratic equation has no real solutions (complex or imaginary solutions).

Now, let's analyze the two equations:

First Equation:

The solutions are −12± √144−120 ÷ 6

Simplify the expression inside the square root:

√ 144 − 120 = 24

This means the discriminant is

D= 144− 4⋅1⋅24 = 144 − 96 = 48.

Since D>0, the first equation will have two distinct real solutions.

Second Equation:

The solutions are −10√100 ± √100−192 ÷ 8

Simplify the expression inside the square root:

√100−192 = √- 92

This means the discriminant is D= 100 − 4⋅1⋅ (−92) = 100 + 368 = 468.

Since D>0, the second equation will also have two distinct real solutions.

So, both equations have real solutions because the discriminants in both cases are greater than zero.

Answer 2

`\cfrac{-12\pm√(144-120)}{6}\implies \cfrac{-12\pm√(24)}{6}\implies \stackrel{~\hfill~ \textit{two real solutions} }{\cfrac{-12\pm2√(6)}{6}\implies \cfrac{-6\pm√(6)}{3}} \\\\[-0.35em] ~\dotfill\\\\ \cfrac{-10\pm√(100-192)}{8}\implies \cfrac{-10\pm√(-92)}{8}\implies \cfrac{-10\pm2√(-23)}{8} \\\\\\ \cfrac{-10\pm2√((-1)(23))}{8}\implies \stackrel{ \textit{two non-real or imaginary solutions} }{\cfrac{-10\pm2\sqrt-{1}\cdot √(23)}{8}\implies \cfrac{-5\pm i√(23)}{4}}`

in essence, if the expression in the square root is negative, we'd end up with two complex or imaginary solutions, if it's positive, then we'd have real solutions, as you can see above why, because √(-1) = i.