+2653 Find all solutions to the equation $x^2 - 11x - 42 = -2x + 10 - x^2 - 5x + 16$.
+135 Sorry, I've just noticed an error with my answer. The starting equation should simplify to 2x^2 - 4x - 68. You can use the same process to find out that the roots are 1 + sqrt(35) and 1 - sqrt(35)
Full process:
\(2x^2-4x-68=0\)
\(x^2-2x-34=0\)
\(x^2-2x+1=35\)
\((x-1)^2 = 35\)
\(x-1 = \pm \sqrt{35}\)
\(x = 1 \pm \sqrt{35}\)
+135 We can first simplify this equation by combining like terms
\(x^2+x^2-11x+2x+5x-42-10-16 = 0\)
\(2x^2+4x-68 = 0\)
\(x^2+4x-34=0\)
Now we can use completing the square to solve this polynomial equation.
\(x^2+4x -34 + 38 = 0 + 38\)
\(x^2+4x+4 = 38\)
\((x+2)^2 = 38\)
\(x + 2 = \pm \sqrt{38}\)
\(x = -2 \pm \sqrt{38}\)
So the 2 solutions are -2 + sqrt(38) and -2 - sqrt(38)
+135 Sorry, I've just noticed an error with my answer. The starting equation should simplify to 2x^2 - 4x - 68. You can use the same process to find out that the roots are 1 + sqrt(35) and 1 - sqrt(35)
Full process:
\(2x^2-4x-68=0\)
\(x^2-2x-34=0\)
\(x^2-2x+1=35\)
\((x-1)^2 = 35\)
\(x-1 = \pm \sqrt{35}\)
\(x = 1 \pm \sqrt{35}\)
