(x+5)⁴+(x+3)⁴=16 how to solve it

y=x+4(1)

Then we have:

(y−1)4+(y+1)4=16

Expanding binomials, odd powers of y cancel, giving

2y4+12y2+2=16

⟹2y4+12y2−14=0

solve this quadratic in y2 for y2 then take square roots on both sides:

y=±−12±122+4⋅2⋅14−−−−−−−−−−−√4−−−−−−−−−−−−−−−−−−−√

substitute into (1):

x=−4±−12±122+4⋅2⋅14−−−−−−−−−−−√4−−−−−−−−−−−−−−−−−−−√

If you evaluate this for all four ± combinations the two real roots turn out to be −5 and −3.

Step-by-step explanation:

answer

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step-by-step explanation:

starting with:    

5 + x - 12 = 2x - 7   

5 + x - 12 - x = 2x - 7 - x (subtract x from both sides)   

5 - 12 = x - 7 (combine and cancel x's from both sides)   

5 - 12 + 7 = x - 7 + 7 (add 7 to both sides)   

0 = x (add numbers together on both sides)       

so in order for the equation to be true, x must be 0.   

trying x = -0.5   

5 + x - 12 = 2x - 7 (original equation)   

5 + -0.5 - 12 = 2(-0.5) - 7 (plug in value of x)   

5 + -0.5 - 12 = -1 - 7 (did multiplication)   

-7.5 = -8 (did addition on both sides)       

since -7.5 is obviously not equal to -8, the value of -0.5 for x is wrong.   

trying x = 0   

5 + x - 12 = 2x - 7 (original equation)   

5 + 0 - 12 = 2(0) - 7 (plug in value of x)   

5 + 0 - 12 = 0 - 7 (did multiplication)   

-7 = -7 (did addition on both sides)       

since -7 equals -7, that demonstrates that the value of 0 for x is correct.   

trying x = 1   

5 + x - 12 = 2x - 7 (original equation)   

5 + 1 - 12 = 2(1) - 7 (plug in value of x)   

5 + 1 - 12 = 2 - 7 (did multiplication)   

-6 = -5 (did addition on both sides)       

since -6 is obviously not equal to -5, the value of 1 for x is wrong.