Question

What are the approximate solutions of 4x2 + 3 = -12x to the nearest hundredth?

Answer 1

Change it into a quadratic equation: ax^2 + bx + c = 0

To do this Add 12x to both sides.

4x^2 + 12x + 3 = 0

a = 4
b = 12
c = 3

Use the Quadratic formula to find the solutions for x.

\[x = \frac{ -b \pm \sqrt{b^2 - 4ac} }{ 2a }\]
\[x = \frac{ -12 \pm \sqrt{12^2 - 4(4)(3)} }{ 2(4) }\]
\[x = \frac{ -12 \pm \sqrt{144 - 48} }{ 8 }\]
\[x = \frac{ -12 \pm \sqrt{96} }{ 8 }\]
\[x = \frac{ -12 \pm 9.798 }{ 8 }\]
\[x = \frac{ -12 + 9.798 }{ 8 }= -0.276\]
\[x = \frac{ -12 - 9.798 }{ 8 }= -2.723\]

So the solutions are:
x = -0.276
and
x = -2.723

Answer 2

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