(x+4)square+(x-4) square=2x(x-5)+8
Answers
Answer:
Rearrange:
Rearrange the equation by subtracting what is to the right of the equal sign from both sides of the equation :
(x+4)^2+(x-4)^2-(2*x*(x-5)+8)=0
Step by step solution :
Step 1 :
Equation at the end of step 1 :
(((x+4)2)+((x-4)2))-(2x•(x-5)+8) = 0
Step 2 :
Step 3 :
Pulling out like terms :
3.1 Pull out like factors :
10x + 24 = 2 • (5x + 12)
Equation at the end of step 3 :
2 • (5x + 12) = 0
Step 4 :
Equations which are never true :
4.1 Solve : 2 = 0
This equation has no solution.
A a non-zero constant never equals zero.
Solving a Single Variable Equation :
4.2 Solve : 5x+12 = 0
Subtract 12 from both sides of the equation :
5x = -12
Divide both sides of the equation by 5:
x = -12/5 = -2.400
solution found :
x = -12/5 = -2.400
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Answer:
Step-by-step explanation:
(x+4)^2 + (x-4)^2 = 2x(x-5)+8
Or 2(x^2+16) = 2x^2 - 10x +8
Or 32 = -10x+8
Or x = -2.4