How do you solve -7sqrt(3x-8)+3=-4x+7?

1 Answer
Feb 26, 2017

You will have to use the quadratic method after simplifying and moving all terms to one side of the equation.

Explanation:

Original Equation:
-7\sqrt(3x-8)+3=-4x+7

Subtract 3 from both sides...
-7\sqrt(3x-8)\cancel(+3)\cancel(-3)=-4x+7-3
-7\sqrt(3x-8)=-4x+4

Divide both sides by -7...
\frac{\cancel(-7)\sqrt(3x-8)}{\cancel(-7)}=(-4x+4)/(-7)
\sqrt(3x-8)=(-4(x-1))/(-7)

Square both sides to get rid of square root...
(\sqrt(3x-8))^2=((-4(x-1))/(-7))^2
3x-8=(16(x-1)^2)/(49)

Multiply both sides by 49...
49(3x-8)=\cancel(49)\cdot(16(x-1)^2)/(\cancel(49))
147x-392=16(x-1)^2

Expand the square (x-1)^2, then simplify...
147x-392=16(x^2-2x+1)
147x-392=16x^2-32x+16

Move all terms to one side and solve the polynomial.
0=16x^2-32x-147x+16+392
0=16x^2-179x+408

Use the quadratic formula to solve for x.
x=(-b+-sqrt(b^2-4ac))/(2a)

x=(179+-sqrt(-179^2-4*16*408))/(2*16)

Simplify.
x=(179+-sqrt(32041-26112))/(32)

x=(179+-sqrt(5929))/(32)

x=(179+-77)/(32)

x=(256)/(32) AND x=(102)/(32)

x=8 AND x=51/16