Question

Find the sum of all real solutions of the equation
x^2−16=3(x−4)^2.

Answer 1

Solve for x over the real numbers:
x^2-16 = 3 (x-4)^2
Write the quadratic polynomial on the right hand side in standard form.
Expand out terms of the right hand side:
x^2-16 = 3 x^2-24 x+48
Move everything to the left hand side.
Subtract 3 x^2-24 x+48 from both sides:
-2 x^2+24 x-64 = 0
Factor the left hand side.
The left hand side factors into a product with three terms:
-2 (x-8) (x-4) = 0
Divide both sides by a constant to simplify the equation.
Divide both sides by -2:
(x-8) (x-4) = 0
Solve each term in the product separately.
Split into two equations:
x-8 = 0 or x-4 = 0
Look at the first equation: Solve for x.
Add 8 to both sides:
x = 8 or x-4 = 0
Look at the second equation: Solve for x.
Add 4 to both sides:
Answer: |
| x = 8 or x = 4

Answer 2

\[\left( a-b \right)^{2}=a ^{2}+b ^{2}-2ab\]

Answer 3

\[ax ^{2}+bx+c=0\]
sum of roots=-b/a

Answer 4

so the answer is 4 ?

Answer 5

let me check

Answer 6

x^2-16=3(x-4)^2
x^2-16=3(x^2-8x+16)
2x^2-24x+64=0
x^2-12x+32=0
sum of roots=-(-12)/1=12