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# Algebra

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Find all solutions to the equation $x^2 - 11x - 42 = -2x + 10 - x^2 - 5x + 16$.

Jun 9, 2024

#2
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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:

$$2x^2-4x-68=0$$

$$x^2-2x-34=0$$

$$x^2-2x+1=35$$

$$(x-1)^2 = 35$$

$$x-1 = \pm \sqrt{35}$$

$$x = 1 \pm \sqrt{35}$$

Jun 9, 2024

#1
+74
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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)

Jun 9, 2024
#2
+74
+1

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:

$$2x^2-4x-68=0$$

$$x^2-2x-34=0$$

$$x^2-2x+1=35$$

$$(x-1)^2 = 35$$

$$x-1 = \pm \sqrt{35}$$

$$x = 1 \pm \sqrt{35}$$

Maxematics  Jun 9, 2024