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

Answers

Answer 1

Answer:

[tex]x=\frac{3}{4},\:x=-1[/tex]

Keys:

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

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

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

[tex]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}\\[/tex]

[tex]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[/tex]

[tex]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}\\[/tex]

solve for x₁

[tex]\frac{-1+7}{2\cdot \:4}[/tex]

[tex]=\frac{6}{2\cdot \:4}[/tex]

[tex]=\frac{6}{8}[/tex]

[tex]= \frac{6\div2}{8\div2}[/tex]

[tex]=\frac{3}{4}[/tex]

solve for x₂

[tex]\frac{-1-7}{2\cdot \:4}[/tex]

[tex]=\frac{-8}{2\cdot \:4}[/tex]

[tex]=\frac{-8}{8}[/tex]

[tex]=-\frac{8}{8}[/tex]

[tex]=-1[/tex]

Hope this helps!


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