Medal! Find the x-intercept of the following quadratic equation.
\[\large x^2 + 5\]
@Nnesha
x-intercept is a point where graph intersect the x-axis when y =0 so \[\large\rm 0=x^2+5\] solve for x and remember (x-intercept is the same thing as solution zeros )so you can use discriminant to find out what kind of solution u r going to get
can i do: \[\large x^2 + 0x + 5 = 0 \rightarrow \frac{ -0 \pm \sqrt{0 - 4(5)} }{ 2 } \rightarrow \frac{ \sqrt{-20} }{ 2 }\] i don't even know if this is correct
that looks good
wouldn't this be no solution?
now as you see there is a negative sign under the root so you're going to get imaginary solution
can*
\(\huge\color{reD}{\rm b^2-4ac}\) `Discriminant` if ` b^2-4ac > 0` then there are 2 real zeros if ` b^2-4ac = 0` then there is one real root if ` b^2-4ac < 0` then you will get two complex roots (no -x-intercept)
real roots means graph will touch the x-axis |dw:1447797949856:dw|
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