Question 1123897
Please help me solve this equation by method of completing the square method: 2x^2+13x-15=0
<pre>{{{matrix(1,3, 2x^2 + 13x - 15, "=", 0)}}}
{{{matrix(1,3, 2(x^2 + (13/2)x - 15/2), "=", 2(0))}}} --- Dividing by 2 to make the coefficient on the x<sup>2</sup>, 1
{{{matrix(1,3, x^2 + (13/2)x - 15/2, "=", 0)}}}
{{{matrix(1,3, x^2 + (13/2)x, "=", 15/2)}}} ------------- Adding {{{15/2}}} to move constant to right of equals sign
{{{matrix(1,3, x^2 + (13/2)x + (13/4)^2, "=", 15/2 + (13/4)^2)}}} ------ Taking {{{1/2}}} of b, squaring it, and then ADDING the result to both sides of equation
{{{matrix(1,3, (x + 13/4)^2, "=", 120/16 + 169/16)}}}
{{{matrix(1,3, (x + 13/4)^2, "=", 289/16)}}}
{{{matrix(1,3, sqrt((x + 13/4)^2), "=", " " +-sqrt(289/16))}}} ------- Taking square root of both sides
{{{matrix(1,3, x + (13/4), "=", " "+- 17/4)}}}
{{{matrix(1,3, x, "=", " " +- 17/4 - 13/4)}}}
{{{system(matrix(1,3, x, "=", 17/4 - 13/4), or, matrix(1,3, x, "=", - 17/4 - 13/4))}}} ========> {{{highlight_green(system(matrix(1,5, x, "=", 4/4, "=", 1), or, matrix(1,7, x, "=", - 30/4, "=", - 7&1/2, or, - 7.5)))}}}
ANY other answer is INCORRECT, so SIMPLY IGNORE!