How do you solve the equation $x^2-3/2x-23/16=0$ by completing the square?

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Answer 1 · 2016-11-13T16:56:09

$x=3/4+sqrt2$ or $x=3/4-sqrt2$

$x^2-3/2x-23/16=0$

completing square of form $(x-a)^2=x^2-2ax+a^2$ , above becomes

$x^2-2xx3/4xx x+(3/4)^2-(3/4)^2-23/16=0$

or $(x-3/4)^2-9/16-23/16=0$ or $(x-3/4)^2-32/16=0$

or $(x-3/4)^2-(sqrt2)^2=0$

and factorizing it using $a^2-b^2=(a+b)(a-b)$ , we get

$(x-3/4-sqrt2)(x-3/4+sqrt2)=0$

Hence, either $x-3/4-sqrt2=0$ i.e. $x=3/4+sqrt2$

or $x-3/4+sqrt2=0$ i.e. $x=3/4-sqrt2$

How do you solve the equation x^2-3/2x-23/16=0x^2-3/2x-23/16=0 by completing the square?