Question

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

Answer 1

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

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

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

$=\frac{-8}{8}$

$=-\frac{8}{8}$

$=-1$

Hope this helps!