Question

y = x^2 + x + 1 has
A)2 rational x-intercepts
B)2 irrational x-intercepts
C)1 rational x-intercept
D)0 x-intercepts

Answer 1

for x-intercepts
solve x^2 + x + 1 =0
\[x=(-1 \pm \sqrt{1-4})/2=(-1 \pm \sqrt{-3})/2\]
so you will have no x-intercepts, becouse roots are complex

Answer 2

Alternatively...

Use the discriminant formula

D = b^2 - 4ac

D = (1)^2 - 4(1)(1) ... Plug in a = 1, b = 1 and c = 1

D = 1 - 4

D = -3

Since the discriminant is negative, this means that there are no real solutions.

So there are no x-intercepts.

Answer 3

what bout for y = 4x^2 + 12x + 9

Answer 4

do the same process again

Answer 5

so i just do B^2-4ac

Answer 6

Same thing, but now a = 4, b = 12 and c = 9

D = b^2 - 4ac

D = (12)^2 - 4(4)(9)

D = 144 - 144

D = 0

Even though we followed the same process, the fact that we got D = 0 tells us that there are 2 identical rational (and real) solutions.

Answer 7

what if its a postive like D=9

Answer 8

Then you get two distinct real solutions

Since 9 is a perfect square, this means that these two solutions are also rational (ie fractions)

Answer 9

thanks guys :)

Answer 10

sure thing