Question

In the following equation, identify the x-intercepts in the graph y = 3x^2 + 19x – 40

Answer 1

factor it out

Answer 2

i got (3x+8)(x-5)
3x+8=0 x-5=0
x=8/3 x=5

is that right?

Answer 3

let's check to see if (3x+8)(x-5) is the correct factorization

(3x+8)(x-5)

3x(x-5)+8(x-5)

3x^2-15x+8x-40

3x^2-7x-40


Since the result above is NOT 3x^2 + 19x – 40 (which is the original expression), this means that 3x^2 + 19x – 40 does NOT factor to (3x+8)(x-5)

Answer 4

oh yeahh. i cant figure it outt

Answer 5

The most direct way (without guessing and checking) is to use the quadratic formula

Answer 6

Are you familiar with it?

Answer 7

ehh not off the top of my head

Answer 8

The quadratic formula is

\[\Large x = \frac{-b\pm\sqrt{b^2-4ac}}{2a}\]

Answer 9

oh okay ik what it is

Answer 10

In the case of \[\Large y = 3x^2 + 19x – 40\]

a = 3, b = 19 and c = -40

Answer 11

i got it! thank youu!!

Answer 12

sure thing

Answer 13

tell me what you get this time

Answer 14

5/3 and 8

Answer 15

close

Answer 16

one of those solutions is off by a sign

Answer 17

-5/3?

Answer 18

no, it's the other one actually, the 8 should be -8

Answer 19

ohhh lol thankss!

Answer 20

here's why

\[\Large x = \frac{-b\pm\sqrt{b^2-4ac}}{2a}\]

\[\Large x = \frac{-(19)\pm\sqrt{(19)^2-4(3)(-40)}}{2(3)}\]

\[\Large x = \frac{-19\pm\sqrt{361-(-480)}}{6}\]

\[\Large x = \frac{-19\pm\sqrt{841}}{6}\]

\[\Large x = \frac{-19+\sqrt{841}}{6} \ \text{or} \ x = \frac{-19-\sqrt{841}}{6}\]

\[\Large x = \frac{-19+29}{6} \ \text{or} \ x = \frac{-19-29}{6}\]

\[\Large x = \frac{10}{6} \ \text{or} \ x = \frac{-48}{6}\]

\[\Large x = \frac{5}{3} \ \text{or} \ x = -8\]

\[\Large x = -8 \ \text{or} \ x = \frac{5}{3}\]

Answer 21

i gotcha!

Answer 22

alright great