Question

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

Answer 1

`x=(3)/(4),\:x=-1`

Keys:

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

• 
  `x_(1,\:2)=(-b\pm √(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.

Explanation:

solving for x₁ and x₂

`4x^2+x=3\\4x^2+x-3=3-3\\4x^2+x-3=0\\x_(1,\:2)=(-1\pm √(1^2-4\cdot 4\left(-3\right)))/(2\cdot 4)\\`

`1^2=1\\=√(1-4\cdot \:4\left(-3\right))\\=√(1+4\cdot \:4\cdot \:3)\\=√(1+48)\\=√(49)\\=√(7^2)\\√(7^2)=7\\=7`

`x_(1,\:2)=(-1\pm \:7)/(2\cdot \:4)\\x_1=(-1+7)/(2\cdot \:4),\:x_2=(-1-7)/(2\cdot \:4)\\`

solve for x₁

`(-1+7)/(2\cdot \:4)`

`=(6)/(2\cdot \:4)`

`=(6)/(8)`

`= (6/2)/(8/2)`

`=(3)/(4)`

solve for x₂

`(-1-7)/(2\cdot \:4)`

`=(-8)/(2\cdot \:4)`

`=(-8)/(8)`

`=-(8)/(8)`

`=-1`

Hope this helps!