How am I supposed to solve x^2 - 6x = 7 by completing the square and get 7 and -1? Like I'm supposed to show work to get that and I'm not getting the right answer



2 years ago Comment

Best answer

This is a quadratic equation so:
[latex] x^{2} [/latex]- 6x = 7
[latex] x^{2} [/latex]-6x-7=0
then find 2 numbers that satisfy [latex] n_{1} [/latex]+[latex] n_{2} [/latex]  = -6 and [latex] n_{1} [/latex]*[latex] n_{2} [/latex] = -7 (where n is one of the numbers)
this works with -1 and 7 (-1 + 7 = 6, -1 * 7 = -7)
[latex] x^{2} [/latex]-6x-7 = (x-1)(x+7)
the awnsers are: -1 and 7


2 years ago Comment

ok so if you have ax^2+bx+c=d to complete the square make sure a=1, if not, dividide everybody by a and since a/a=1, will now equal 1 then move c to other side the take 1/2 of b and square it then add this new number and add it to both sides then factor perfect sqare then take square root of both sides then find x x^2-6x=7 a=1 and c is already on other side take 1/2 of b and square it -6/2=-3 (-3)^2=9 add that to both sides x^2-6+9=7+9 factor perfect square (remember that middle sign is the same sign as b) (x-3)^2=16 squaer root both sides and remember to get the positive and negative root x-3=-4 or 4 2 equations x-3=4 x-3=-4 add 3 to both sides x=7 x=-1 tada