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

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Answer 1 · 2016-08-13T04:06:40

$x=2/3-sqrt2$ or $x=2/3+sqrt2$

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

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

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

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

$(3x-2)^2-(sqrt18)^2=0$

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

$(3x-2+sqrt18)(3x-2-sqrt18)=0$

i.e. either $3x-2+sqrt18=0$ or $3x-2-sqrt18=0$ .

Now as $sqrt18=3sqrt2$

either $x=2/3-sqrt2$ or $x=2/3+sqrt2$

How do you solve 9x^2-12x-14=09x^2-12x-14=0 by completing the square?