rewrite in standard form -2x^2+3x+7

Question

rewrite in standard form -2x^2+3x+7

Answer 1

Recall that the standard form of a quadratic equation is as follows:   a(x - h)² + k

     -2x² + 3x + 7          separate the first 2 terms from the last term by parentheses

     (-2x² + 3x) + 7        

factor out the coefficient of the 1st term (a) from the first 2 terms which are inside the parentheses

     -2(x² - (3/2)x) + 7

Complete the square inside the parentheses:

     x² - (3/2)x          ==>    (b/2)² = ((-3/2)/2)² = (-3/4)² = 9/16

Add and subtract the term that completes the square inside the parentheses then multiply the term that is subtracted by a (i.e., -2), which was factored out in the beginning, then move this term to the outside.

     -2(x² - (3/2)x + (9/16) - (9/16)) + 7

     -2(x² - (3/2)x + (9/16)) + 7 - (9/16)·(-2)

     -2(x² - (3/2)x + (9/16)) + 7 - (-2·9/16)

     -2(x² - (3/2)x + (9/16)) + 7 + (18/16)

==>  7 + (18/16) = (7·16/16) + (18/16) = (112/16) + (18/16) = (112+18)/16 = 130/16 = 65/8

     -2(x² - (3/2)x + (9/16)) + 65/8

Since we completed the square for the equation inside the parentheses in the beginning, the equation inside the parentheses is now a perfect square:

     -2(x - (3/4))(x - (3/4)) + 65/8

     -2(x - (3/4))² + 65/8