Question

5. -7x^2+6x+3=0
and
-8x^2-8x-2=0

for each one how many real number soloutions does the equation have?

Answer 1

use discriminant: b^2-4ac in quadratic equation ax^2+bx+c=0
if discriminant > 0, 2 solutions
if =, 1 solution
if < 0, no solutions

Answer 2

I dont understand

Answer 3

so for the first one the discriminant is (6)^2-4(-7)(3) = 36+84 = 120
Therefore, two solutions

Answer 4

Do you get it?

Answer 5

\[\begin{array}{l}{x_{1/2}} = \frac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\\\\\Delta = {b^2} - 4ac\\\Delta > 0\\\Delta = 0\\\Delta < 0\end{array}\]

Answer 6

I can explain further if necessary

Answer 7

please do

Answer 8

OK then...

Answer 9

Here's the graph of y = -7x^2+6x+3

From the graph we can see that -7x^2+6x+3 = 0 is true at 2 points
Therefore it has 2 solutions

Answer 10

what about the second one?

Answer 11

-8x^2-8x-2=0

Let's graph it

Answer 12

\[\LARGE -7x^2+6x+3=0\]
now use the quadratic formula:
\[\LARGE {x_{1/2}} = \frac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\]

\[\LARGE \begin{array}{l}{x_{1/2}} = \frac{{ - 6 \pm \sqrt {36 - 4( - 7)(3)} }}{{2( - 7)}} \ \ \\ \\ \\{x_{1/2}} = \frac{{ - 6 \pm \sqrt {36 + 84} }}{{ - 14}}\\{x_{1/2}} = \frac{{ - 6 \pm \sqrt {120} }}{{ - 14}}\end{array}\]
so...? (the first one)

Answer 13

We can see it only touches the x axis (where y = 0) once, so it has one solution.

Using the discriminant: (-8)^2 - 4(-8)(-2) = 0. Therefore only one solution

Answer 14

thank you so much!!

Answer 15

No probs