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inna [77]
2 years ago
12

Find the solution set. 4x^2+x=3

Mathematics
1 answer:
Zina [86]2 years ago
3 0

Answer:

x=\frac{3}{4},\:x=-1

Keys:

For this problem, you need the quadratic formula(listed below).

  • x_{1,\:2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}
  • 1^a=1
  • \sqrt[n]{a}^n=a

When you see ± in a quadratic equation, you must know there is going to be at least 2 solutions.

Step-by-step explanation:

solving for x₁ and x₂

4x^2+x=3\\4x^2+x-3=3-3\\4x^2+x-3=0\\x_{1,\:2}=\frac{-1\pm \sqrt{1^2-4\cdot 4\left(-3\right)}}{2\cdot 4}\\

1^2=1\\=\sqrt{1-4\cdot \:4\left(-3\right)}\\=\sqrt{1+4\cdot \:4\cdot \:3}\\=\sqrt{1+48}\\=\sqrt{49}\\=\sqrt{7^2}\\\sqrt{7^2}=7\\=7

x_{1,\:2}=\frac{-1\pm \:7}{2\cdot \:4}\\x_1=\frac{-1+7}{2\cdot \:4},\:x_2=\frac{-1-7}{2\cdot \:4}\\

solve for x₁

\frac{-1+7}{2\cdot \:4}

=\frac{6}{2\cdot \:4}

=\frac{6}{8}

= \frac{6\div2}{8\div2}

=\frac{3}{4}

solve for x₂

\frac{-1-7}{2\cdot \:4}

=\frac{-8}{2\cdot \:4}

=\frac{-8}{8}

=-\frac{8}{8}

=-1

<em>Hope this helps!</em>

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