Question

the values of x for which the graphs of y=x+2,y^2=4x intersect are:
a. -2 and 2
b. -2
c. 2
d. 0
e. none of these

Answer 1

\[\large\rm y=x+2,\qquad\qquad y^2=4x\]We have a couple ways to approach this...
Hmm, let's try solving each equation for x.

Answer 2

In the first equation, subtract 2 gives us,\[\large\rm y-2=\color{orangered}{x}\]And in the second equation, dividing by 4 gives us,\[\large\rm \frac{y^2}{4}=\color{orangered}{x}\]They will intersect when these x's are the same, which means the equations are the same.\[\large\rm y-2=\frac{y^2}{4}\]

Answer 3

Multiply by 4,\[\large\rm 4y-8=y^2\]Move everything to one side,\[\large\rm y^2-4y+8=0\]

Answer 4

It looks like this `is not` going to factor nicely.
Try using your quadratic formula:\[\large\rm y=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\]Plug in the values, what does it tell you? :)

Answer 5

so its b? @zepdrix

Answer 6

Hmm +_+
Well, we're trying to find our y-coordinate,
and we can then use that to find the x-coordinate of intersection.

What did you get for the y-value?
Use the formula.

Answer 7

-2 and 0 @zepdrix

Answer 8

\(y=x+2\)
\(y^2=4x\)

From above eqn
\(y^2=4(y-2)\)
\(y^2-4y+8=0\)

solve eqn either by quadratic formula or factorization.

Answer 9

\[y=\frac{4 \pm \sqrt{16-32}}{2}\]and \[ x=y-2\]

Answer 10

so im right @jiteshmeghwal9

Answer 11

\[y=\frac{4 \pm 4 \iota}{2}=2\pm2 \iota\]

\[x=y-2=2\pm2 \iota -2\]\[x= \pm 2 \iota\]