Question

What are the approximate solutions of 4x^2 + 3 = −12x to the nearest hundredth?

x ≈ −3.23 and x ≈ 0.23

x ≈ −2.72 and x ≈ −0.28

x ≈ 0.28 and x ≈ 2.72

x ≈ −0.23 and x ≈ 3.23

Answer 1

I dont get this AT ALL :/

Answer 2

Rearrange to form a quadratic:
4x^2 + 3 = -12
4x^2 + 3 + 12x = 0
4x^2 + 12x + 3 = 0

Then solve using the quadratic equation, where the quadratic is of the form ax^2 + bx + c = 0:
\[\frac{-b\pm\sqrt{b^2-4ac}}{2a}\]

Answer 3

the equation is

\[4x^2 + 12x + 3 = 0\]

you need the general quadratic formula with

a = 4, b = 12 and c = 3

\[x = \frac{-12\pm \sqrt{12^2 - 4 \times4 \times3}}{2\times4}\]

the solutions are

\[x = \frac{ -12 p \sqrt{96}}{8}...and...x = \frac{-12 - \sqrt{96}}{8}\]

just evaluate for your answers

Answer 4

oops

\[x = \frac{-12 + \sqrt{96}}{8}\]

Answer 5

so just evaluate from there and thats the answer?

Answer 6

thats correct.... givwn both values of x are negative... it should be easy to find the solutions in
your choices.