Question

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

Answer 1

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`