Question

I WILL GIVE 50 POINTS AND BRAINLIEST!!! HARD ALGEBRA I PORBLEM!!!

Use the discriminant to determine the number and type of solutions to the quadratic equation and solve the questions below. Show all work for full credit.

Use the discriminant to determine the number of solutions and types of solutions for the quadratic equation, below.


a^2+8a=13


A: Discriminant:

B: Number of Solutions:

C: Type of solutions (real or imaginary):

D: Type of solutions (rational or irrational):

If you spam the answers with any non-answers, I WILL report you.

Answer 1

Answer:

The discriminant is the (b^2-4ac) part of the quadratic formula...in this case:

64-48=16

So there will be two real rational solutions.

The rule:

If discriminant<0, there are no real solutions (although there are two imaginary solutions)

If discriminant=0, there is only one real solution.

If discriminant>0, there are two real solutions.

Step-by-step explanation:Sorry bout all these people out here putting fake answers. have a great day!

Answer 2

Answer:

A: $discriminant=\sqrt{116}$

B: 2 solutions

C: 2 Real Solutions

D: Irrational

Step-by-step explanation:

You're given $a^2+8a=13$.

I'll rewrite it in term of x as to not be confused with the other letters I use.

$x^2+8x-13=0$

A:

The quadratic equation is

$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

The discriminant is the root part of the quadratic equation.

$discriminant=\sqrt{b^2-4ac}$

You're given

$x^2+8x-13=0$

so your a=1,  b=8, and c=-13. Plug it in and solve.

$discriminant=\sqrt{(8)^2-4(-13)(1)}$

$discriminant=\sqrt{116}$

B:

If you plug your discriminant back into your quadratic formula, you'll see that it'll give you two solutions.

$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

$x=\frac{-8\pm\sqrt{116}}{2(1)}$

$x=\frac{8}{2}\pm\frac{\sqrt{116}}{2}$

$x=4\pm\frac{\sqrt{116}}{2}$   --->  $x=4+\frac{\sqrt{116}}{2}$,     $x=-\frac{\sqrt{116}}{2}$

Simply, if the discriminant is a positive value then the ± will exist and give you 2 solutions.

The other guy that answered mentioned the shortcut to finding this, but I think it's better to know the reason why those shortcut works.

Note: There's two other possibility you might see. You can ignore this part if you want.

1.) You have a negative discriminant. (negative square root)

$x=\frac{-b\pm\sqrt{-number}}{2a}$

this will give you imaginary solution because squareroot of negative numbers give you an imaginary value. $\sqrt{-1}=i$

2.) Your discriminant cancels out itself. i.e. your discriminant=0.

$x=\frac{-b\pm\sqrt{0}}{2a}=\frac{-b}{2a}$

In this case, the ±√(...) is gone so you'll only have 1 solution.

C:

Positive squareroot is real value.  $\sqrt{+number}=real$

Negative squareroot is imaginary.  $\sqrt{-number} =i$

The 116 in $\sqrt{116}$ is positive so it's real.

D:

Rational numbers are numbers that could be represented as a simple fraction.

$\sqrt{116}=10.7703229...$  and cannot be represented in a form of a simple fraction, therefore it is irrational.