Question

Did I get this right? How many real-number solutions does the equation have?

7x^2 + 8x + 5 = 0

Answer 1

one solution
two solutions**
no solutions
infinitely many solutions

Answer 2

what did you get?

Answer 3

oh nvm, one sec while I check

Answer 4

no that's not correct

Answer 5

the discriminant is

D = b^2 - 4ac

Answer 6

This determines what type and how many solutions you'll have

Answer 7

D = b^2 - 4ac

D = 8^2 - 4(7)(5)

D = 64 - 140

D = -76

because the discriminant is negative, you'll have no real solutions

Answer 8

The solutions instead will be complex or imaginary (if you haven't heard of either term, then don't worry)

Answer 9

Thank you :)

Answer 10

yw

Answer 11

@jim_thompson5910 Could you help with one more?

Answer 12

A model rocket is launched from a roof into a large field. The path of the rocket can be modeled by the equation y = 0.04x+ 8.3x + 4.3, where x is the horizontal distance, in meters, from the starting point on the roof and y is the height, in meters, of the rocket above the ground. How far horizontally from its starting point will the rocket land?
208.02 m
416.03 m
0.52 m
208.19 m

Answer 13

is there an x^2 in there?

Answer 14

y = 0.04x^2+ 8.3x + 4.3 Yes sorry

Answer 15

and is the leading coefficient negative?

Answer 16

Yeah, woops

Answer 17

ok so you you'll need to find the roots of y = -0.04x^2 + 8.3x + 4.3

Answer 18

the best way to do so is to use the quadratic formula

Answer 19

have you heard of that formula before?

Answer 20

So that's ax+b+c ? Not sure if that's right.

Answer 21

The quadratic formula is

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

Answer 22

you would plug a = -0.04, b = 8.3 and c = 4.3 in to get

\[\Large x = \frac{-(8.3)\pm\sqrt{(8.3)^2-4(-0.04)(4.3)}}{2(-0.04)}\]

Answer 23

I'm still confused..

Answer 24

From this point onward, it's just a matter of using a calculator to evaluate and simplify

\[\Large x = \frac{-8.3\pm\sqrt{68.89-(-0.688)}}{-0.08}\]


\[\Large x = \frac{-8.3\pm\sqrt{68.89+0.688}}{-0.08}\]


\[\Large x = \frac{-8.3\pm\sqrt{69.578}}{-0.08}\]


\[\Large x = \frac{-8.3+\sqrt{69.578}}{-0.08} \ \text{or} \ x = \frac{-8.3-\sqrt{69.578}}{-0.08}\]


\[\Large x = \frac{-8.3+\sqrt{69.578}}{-0.08} \ \text{or} \ x = \frac{-8.3-\sqrt{69.578}}{-0.08}\]


\[\Large x = -.5167852195 \ \text{or} \ x = 208.0167852\]


\[\Large x \approx -0.516785 \ \text{or} \ x \approx 208.016785\]

Answer 25

it's a bit of a mess, but in the end, the two roots are approximately

x = -0.516785 or x = 208.016785

Answer 26

Is it a?

Answer 27

ignore the negative root and focus only on the positive root

Answer 28

yes it is

Answer 29

going from x = 0 to x = 208.016785 is roughly 208.02 ft

Answer 30

Alright. Thanks again!

Answer 31

sure thing