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

User Fielding
by
3.0k points

1 Answer

19 votes
19 votes

Answer:


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!

User Cyraxjoe
by
2.8k points
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