How do you solve for x: $(x - 3) (x - 12) = (2 x + 5) (x - 6)$ $(x-3)(x-12)=(2x+5)(x-6)$ ?

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How do you solve for x: $(x - 3) (x - 12) = (2 x + 5) (x - 6)$ $(x-3)(x-12)=(2x+5)(x-6)$ ?

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Answer 1 · 2017-06-17T20:19:10

$x = - 4 \pm \sqrt{82}$ $x = -4+-sqrt82$

Explanation:

First, use the FOIL method to expand both sides:

$x^{2} - 12 x - 3 x + 36 = 2 x^{2} - 12 x + 5 x - 30$ $x^2 - 12x - 3x + 36 = 2x^2 - 12x + 5x - 30$

$x^{2} - 15 x + 36 = 2 x^{2} - 7 x - 30$ $x^2 - 15x + 36 = 2x^2 - 7x - 30$

Now, move all the left terms to the right side by subtracting.

$0 = x^{2} + 8 x - 66$ $0 = x^2 + 8x - 66$

Hmm... this can't be factored. But, we can complete the square to solve it. We know that ${(x + 4)}^{2} = x^{2} + 8 x + 16$ $(x+4)^2 = x^2 + 8x + 16$ , so:

$0 = x^{2} + 8 x + 16 - 16 - 66$ $0 = x^2 + 8x + color(blue)16 - color(blue)16 - 66$

$0 = {(x + 4)}^{2} - 82$ $0 = (x+4)^2 - 82$

Now, we can solve this using algebra:

$82 = {(x + 4)}^{2}$ $82 = (x+4)^2$

$\pm \sqrt{82} = x + 4$ $+-sqrt82 = x+4$

$- 4 \pm \sqrt{82} = x$ $-4+- sqrt82 = x$

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How do you solve for x: $(x - 3) (x - 12) = (2 x + 5) (x - 6)$ $(x-3)(x-12)=(2x+5)(x-6)$ ?

1 Answer

Explanation:

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How do you solve for x: (x-3)(x-12)=(2x+5)(x-6)(x−3)(x−12)=(2x+5)(x−6) (x-3)(x-12)=(2x+5)(x-6)?

1 Answer

Explanation:

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How do you solve for x: $(x - 3) (x - 12) = (2 x + 5) (x - 6)$ $(x-3)(x-12)=(2x+5)(x-6)$ ?