That one looks very difficuly
Vertex = (1/2,5/2) x= 1 + or - square root of 5-2y all over 2
Recall that given the equation of the second degree (or quadratic)
ax ^ 2 + bx + c
Its solutions are:
x = (- b +/- root (b ^ 2-4ac)) / 2a
discriminating:
d = root (b ^ 2-4ac)
If d> 0, then the two roots are real (the radicand of the formula is positive).
If d = 0, then the root of the formula is 0 and, therefore, there is only one solution that is real and of multiplicity 2 (it is a double root).
If d <0, then the two roots are complex and, in addition, one is the conjugate of the other. That is, if one solution is x1 = a + bi, then the other solution is x2 = a-bi (we are assuming that a, b, c are real).
One solution:
A cut point with the x axis
Two solutions:
Two cutting points with the x axis.
Complex solutions:
Does not cut to the x axis
Answer:
Exact form : x = 4 ± √14
Decimal form : x 7.74165738..., 0.25834261...
Step-by-step explanation:
I hope this helps you out. ☺
