How do you solve by completing the square for $2x^2+7x - 15 = 0$ ?

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Answer 1 · 2015-03-30T17:47:23

Group the two variable terms together, then factor out the number in front of $x^2$

$(2x^2+7x)-15=0$

$2(x^2+7/2 x+ color(white)"leave space") -15=0$

Take $1/2$ of the coefficient of $x$ , square that and add and subtract inside the parentheses:

$1/2 " of " 7/2 = 7/4$ Squaring gives us $49/16$ , so we write:

$2(x^2+7/2 x+ 49/16 - 49/16) -15=0$ Keep the positive $49/16$ inside the parentheses (we need it for the perfect square)

Write:
$2(x^2+7/2 x+ 49/16) - 2(49/16) -15=0$

Factor the square and simplify the rest:

$2(x + 7/4)^2 - 49/8 - 120/8=0$

$2(x + 7/4)^2 - 169/8=0$

$2(x + 7/4)^2 = 169/8$

$(x + 7/4)^2 = 169/16$

$x + 7/4 = +- sqrt(169/16)$

$x + 7/4 = +- 13/4$

$x = -7/4 +- 13/4$

$-7/4 + 13/4 = 6/4=3/2$ and $-7/4 - 13/4 = -20/4 = -5$

The solutions are $3/2$ , $5$

How do you solve by completing the square for 2x^2+7x - 15 = 02x^2+7x - 15 = 0?