Question

use the quadratic formula to find any x-intercepts on the graph of the equation y=x^2-6x-4

Answer 1

for x-intercept put y=0 (eq of x-axis)
i.e x^2-6x-4=0
x= -3 +- (sqrt(52))/2 =-3+-sqrt(13) (the reqd x-intercepts)

Answer 2

can you write it out in equation form...i'm confused here

Answer 3

@matricked

Answer 4

oops
it is x= 3 +- (sqrt(52))/2 =-3+-sqrt(13) (the reqd x-intercepts)

Answer 5

x^2-6x+9-9-4=0
or(x-3)^2=13
or ((x-3)^2-(sqrt(13))^2)=0
or (x-3+sqrt(13))(x-3-sqrt(13))=0
hence x=3-sqrt(13) or x=3+sqrt(13)

Answer 6

\[y=x^2-6x-4\]\[x =\frac{ -B \pm \sqrt{B^2-4AC} }{ 2A } <- Quadratic formula\]\[x=\frac{ -(-6) \pm \sqrt{(-6)^2-4(1)(-4)} }{ 2(1) }\] \[x=\frac{ 6 \pm \sqrt{36+16} }{ 2 }\]\[x=\frac{ 6 \pm \sqrt{52} }{ 2 }\]

Answer 7

Simplify to get:\[x=3-\sqrt{13}\]and\[x=3+\sqrt{13}\]