This quadratic equation which will have two solutions can be solved by completing the square or by using the quadratic equation formula.
Completing the square:
x2+18x+4 = 0
(x+9)2+4 = 0
(x+9)2+4-81 = 0
(x+9)2 = 77
x+9 = + or - the square root of 77
x = -9 + or - the square root of 77
If you're not too sure about the procedure of completing the square your maths tutor should be familiar with it.
x2-18x+81 = (x-9)(x-9) when factored
2x2+18x = 20 2x2+18x-20 = 0 x2+9x-10 = 0 (x+10)(x-1) = 0 x = -10 or x = 1
Y = X2 + 18X + 52 set to 0 X2 + 18X + 52 = 0 subtract 52 from each side X2 + 18X = - 52 now, halve the coefficient of the variable term (18), square it and add it to both sides X2 + 18X + 81 = - 52 + 81 now, factor on left and gather terms on right (X + 9)2 = 29 (X + 9)2 - 29 = 0 ---------------------------vertex form
x2=18x-80 (im guessing you are trying to find x) x2-18x+80=0 (x-10)(x-8)=0 Therefore: x = 10, 8
The vertex is (-9, -62).
x2 + 18x - 50 does not have rational factors.
x=4 and x =7Improved Answer:-x2-18x-40 = (x+2)(x-20) when factored
Yes!
x2-18x+81 = (x-9)(x-9) when factored
2x2+18x = 20 2x2+18x-20 = 0 x2+9x-10 = 0 (x+10)(x-1) = 0 x = -10 or x = 1
x2 plus 3x plus 3 is equals to 0 can be solved by getting the roots.
Y = X2 + 18X + 52 set to 0 X2 + 18X + 52 = 0 subtract 52 from each side X2 + 18X = - 52 now, halve the coefficient of the variable term (18), square it and add it to both sides X2 + 18X + 81 = - 52 + 81 now, factor on left and gather terms on right (X + 9)2 = 29 (X + 9)2 - 29 = 0 ---------------------------vertex form
x2=18x-80 (im guessing you are trying to find x) x2-18x+80=0 (x-10)(x-8)=0 Therefore: x = 10, 8
The vertex is (-9, -62).
81. To complete the square, halve the coefficient of the x term (18/2 = 9) and add the square of this (92 = 81) to both sides: x2 + 18x = -13 ⇒ x2 + 18x + 81 = -13 + 81 ⇒ (x + 9)2 = 68
(2x)ysquared
9x2 - 18x + 99(x2 - 2x + 1)==========