How do you solve #-3x^2-12=14x# by completing the square?

Question

Julia Adams · Accepted Answer

#x~~ -1.13 or x~~ -3.54# #-3x^2-12=14x or 3x^2+14x+12 =0 or 3(x^2+14/3x)+12# or # 3(x^2+14/3x +(7/3)^2)-(3*49/9)+12 =0# #3(x+7/3)^2-49/3+12 =0 # or #3(x+7/3)^2-13/3 =0 or 3(x+7/3)^2=13/3 # or #(x+7/3)^2=13/9 or (x+7/3) = +-sqrt(13)/3#. Either #x= -7/3+sqrt13/3 or x= -7/3-sqrt13/3# or #x= (sqrt13-7)/3 or x= -(sqrt13+7)/3# or #x~~ -1.13(2dp) or x~~ -3.54(2dp)# [Ans]

Eva Adamson · Answer

undefined To solve the equation $ -3x^2 - 12 = 14x $ by completing the square, follow these steps:

1. Move all terms to one side of the equation to set it equal to zero: $ -3x^2 - 14x - 12 = 0 $.

2. Divide the coefficient of $ x^2 $ by 2 and square the result: $ \left(\frac{-14}{2}\right)^2 = 49 $.

3. Add and subtract the result obtained in step 2 inside the parentheses: $ -3x^2 - 14x + 49 - 49 - 12 = 0 $.

4. Rewrite the equation by grouping the perfect square trinomial: $ -3(x^2 + \frac{14}{3}x + 49) - 49 - 12 = 0 $.

5. Factor the perfect square trinomial: $ -3(x + \frac{7}{3})^2 - 61 = 0 $.

6. Add 61 to both sides of the equation: $ -3(x + \frac{7}{3})^2 = 61 $.

7. Divide both sides by -3: $ (x + \frac{7}{3})^2 = -\frac{61}{3} $.

8. Take the square root of both sides: $ x + \frac{7}{3} = \pm\sqrt{-\frac{61}{3}} $.

9. Subtract $\frac{7}{3}$ from both sides: $ x = -\frac{7}{3} \pm \sqrt{-\frac{61}{3}} $.

10. Simplify the square root if necessary.