How do you solve #3s^2-26x+2=5s^2+1# by completing the square?

Question

Abigail Allen · Accepted Answer

#color(brown)(x = (1/2)(3sqrt19 - 13), -(1/2) (3sqrt19 + 13)# #3x^2 - 26x+ 2 = 5x^2 + 1# #5x^2 - 3x^2 + 26x = 1# #2(x^2 + (2 * x * (13/2)) = 1# #x^2 + 2x * (13/2) = 1/2# Adding # (13/2)^2 # to both sides, #x^2 + 2x (13/2) + (13/2)^2 = 1/2 + (13/2)^2 = (171/4)# #(x + 13/2)^2 = (sqrt(171/4))^2# #x + 13/2 = +- sqrt(171/4)# #x = +- sqrt(171/4) - 13/2# #color(brown)(x = (1/2)(3sqrt19 - 13), -(1/2) (3sqrt19 + 13)#

Nathan Axel · Answer

undefined To solve the equation 3s^2 - 26s + 2 = 5s^2 + 1 by completing the square, follow these steps:

1. Move all terms to one side to set the equation equal to zero: 3s^2 - 26s + 2 - 5s^2 - 1 = 0.
2. Combine like terms: (3s^2 - 5s^2) - 26s + (2 - 1) = 0.
3. Simplify: -2s^2 - 26s + 1 = 0.
4. Divide the entire equation by the coefficient of the squared term to make the coefficient 1: -2(s^2 + (26/2s)) + 1 = 0.
5. Complete the square by adding and subtracting the square of half the coefficient of the linear term: -2(s^2 + 13s + (13/2)^2 - (13/2)^2) + 1 = 0.
6. Simplify the expression inside the parentheses: -2(s^2 + 13s + 169/4 - 169/4) + 1 = 0.
7. Simplify further: -2((s + 13/2)^2 - 169/4) + 1 = 0.
8. Distribute the -2: -2(s + 13/2)^2 + 169/2 + 1 = 0.
9. Combine constants: -2(s + 13/2)^2 + 169/2 + 2/2 = 0.
10. Combine constants further: -2(s + 13/2)^2 + 171/2 = 0.
11. Move constant to the other side: -2(s + 13/2)^2 = -171/2.
12. Divide both sides by -2: (s + 13/2)^2 = 171/4.
13. Take the square root of both sides: s + 13/2 = ±√(171/4).
14. Solve for s: s = -13/2 ± √(171/4).