How do you solve $9 x^{2} - 12 x - 14 = 0$ $9x^2-12x-14=0$ by completing the square?

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How do you solve $9 x^{2} - 12 x - 14 = 0$ $9x^2-12x-14=0$ by completing the square?

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Answer 1 · 2016-08-13T04:06:40

$x = \frac{2}{3} - \sqrt{2}$ $x=2/3-sqrt2$ or $x = \frac{2}{3} + \sqrt{2}$ $x=2/3+sqrt2$

Explanation:

In $9 x^{2} - 12 x - 14 = 0$ $9x^2-12x-14=0$ , while $9 x^{2} = {(3 x)}^{2}$ $9x^2=(3x)^2$ , to complete the square, recall the identity ${(x \pm a)}^{2} = x^{2} \pm 2 a x + a^{2}$ $(x+-a)^2=x^2+-2ax+a^2$ .

As $- 12 x = - 2 \times (3 x) \times 2$ $-12x=-2×(3x)×2$ , we need to add $2^{2}$ $2^2$ to make it complete square.

Hence, $9 x^{2} - 12 x - 14 = 0$ $9x^2-12x-14=0$ can be written as $({(3 x)}^{2} - 2 \times (3 x) \times 2 + 2^{2}) - 4 - 14 = 0$ $((3x)^2-2×(3x)×2+2^2)-4-14=0$ or

${(3 x - 2)}^{2} - 18 = 0$ $(3x-2)^2-18=0$ , which is equivalent to

${(3 x - 2)}^{2} - {(\sqrt{18})}^{2} = 0$ $(3x-2)^2-(sqrt18)^2=0$

and using identity ${(a - b)}^{2} = (a + b) (a - b)$ $(a-b)^2=(a+b)(a-b)$ we can write the equation as

$(3 x - 2 + \sqrt{18}) (3 x - 2 - \sqrt{18}) = 0$ $(3x-2+sqrt18)(3x-2-sqrt18)=0$

i.e. either $3 x - 2 + \sqrt{18} = 0$ $3x-2+sqrt18=0$ or $3 x - 2 - \sqrt{18} = 0$ $3x-2-sqrt18=0$ .

Now as $\sqrt{18} = 3 \sqrt{2}$ $sqrt18=3sqrt2$

either $x = \frac{2}{3} - \sqrt{2}$ $x=2/3-sqrt2$ or $x = \frac{2}{3} + \sqrt{2}$ $x=2/3+sqrt2$

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How do you solve $9 x^{2} - 12 x - 14 = 0$ $9x^2-12x-14=0$ by completing the square?

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Explanation:

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How do you solve $9 x^{2} - 12 x - 14 = 0$ $9x^2-12x-14=0$ by completing the square?