How do you solve $2 x^{2} - 3 x - 14 = 0$ $2x^2-3x-14=0$ by completing the square?

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How do you solve $2 x^{2} - 3 x - 14 = 0$ $2x^2-3x-14=0$ by completing the square?

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Answer 1 · 2016-12-22T01:05:00

$x = \frac{7}{2}$ $x=7/2" "$ or $x = - 2$ $" "x=-2$

Explanation:

The difference of squares identity can be written:

$A^{2} - B^{2} = (A - B) (A + B)$ $A^2-B^2 = (A-B)(A+B)$

We will use this with $A = (4 x - 3)$ $A=(4x-3)$ and $B = 11$ $B=11$ later.

$color(white)()$
Given:

$2 x^{2} - 3 x - 14 = 0$ $2x^2-3x-14 = 0$

To avoid fractions as much as possible, let us premultiply this quadratic by $8 = 2 \cdot 2^{2}$ $8 = 2*2^2$ . One factor of $2$ $2$ makes the leading term a perfect square, then the other $2^{2}$ $2^2$ deals with the $2$ $2$ denominator in " $\frac{b}{2 a}$ $b/(2a)$ " which would otherwise cause us to have to work with $\frac{1}{2}$ $1/2$ 's.

So:

$0 = 8 (2 x^{2} - 3 x - 14)$ $0 = 8(2x^2-3x-14)$

$0 = 16 x^{2} - 24 x - 112$ $color(white)(0) = 16x^2-24x-112$

$0 = {(4 x)}^{2} - 2 (4 x) (3) + 9 - 121$ $color(white)(0) = (4x)^2-2(4x)(3)+9-121$

$0 = {(4 x - 3)}^{2} - 11^{2}$ $color(white)(0) = (4x-3)^2-11^2$

$0 = ((4 x - 3) - 11) ((4 x - 3) + 11)$ $color(white)(0) = ((4x-3)-11)((4x-3)+11)$

$0 = (4 x - 14) (4 x + 8)$ $color(white)(0) = (4x-14)(4x+8)$

$0 = (2 (2 x - 7)) (4 (x + 2))$ $color(white)(0) = (2(2x-7))(4(x+2))$

$0 = 8 (2 x - 7) (x + 2)$ $color(white)(0) = 8(2x-7)(x+2)$

Hence:

$x = \frac{7}{2}$ $x=7/2" "$ or $x = - 2$ $" "x=-2$

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How do you solve $2 x^{2} - 3 x - 14 = 0$ $2x^2-3x-14=0$ by completing the square?

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Explanation:

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How do you solve $2 x^{2} - 3 x - 14 = 0$ $2x^2-3x-14=0$ by completing the square?