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# algebra

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Solve in real numbers: $$x^2 + 7x - 5 = 5 \sqrt{x^3 - 1}$$

May 17, 2020

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Solve for x:
x^2+7 x-5==5 Sqrt[x^3-1]

x^2+7 x-5==5 Sqrt[x^3-1] is equivalent to 5 Sqrt[x^3-1]==x^2+7 x-5:
5 Sqrt[x^3-1]==x^2+7 x-5

Raise both sides to the power of two:
25 (x^3-1)==(x^2+7 x-5)^2

Expand out terms of the left hand side:
25 x^3-25==(x^2+7 x-5)^2

Expand out terms of the right hand side:
25 x^3-25==x^4+14 x^3+39 x^2-70 x+25

Subtract x^4+14 x^3+39 x^2-70 x+25 from both sides:
-x^4+11 x^3-39 x^2+70 x-50==0

The left hand side factors into a product with three terms:
-(x^2-8 x+10) (x^2-3 x+5)==0

Multiply both sides by -1:
(x^2-8 x+10) (x^2-3 x+5)==0

Split into two equations:
x^2-8 x+10==0 or x^2-3 x+5==0

Subtract 10 from both sides:
x^2-8 x==-10 or x^2-3 x+5==0

x^2-8 x+16==6 or x^2-3 x+5==0

Write the left hand side as a square:
(x-4)^2==6 or x^2-3 x+5==0

Take the square root of both sides:
x-4==Sqrt or x-4==-Sqrt or x^2-3 x+5==0

x==4+Sqrt or x-4==-Sqrt or x^2-3 x+5==0

x==4+Sqrt or x==4-Sqrt or x^2-3 x+5==0

Subtract 5 from both sides:
x==4+Sqrt or x==4-Sqrt or x^2-3 x==-5

x==4+Sqrt or x==4-Sqrt or x^2-3 x+9/4==-(11/4)

Write the left hand side as a square:
x==4+Sqrt or x==4-Sqrt or (x-3/2)^2==-(11/4)

Take the square root of both sides:
x==4+Sqrt or x==4-Sqrt or x-3/2==(I Sqrt)/2 or x-3/2==-((I Sqrt)/2)  x =  -(√6) + 4  x = (sqrt6) + 4  