Question

what is the product of the 2 solutions of the equation x^2+3x-21=0

Answer 1

imagine alpha and beta is two solutions of the equation ax^2 + bx + x = 0 so:

alpha = {-b + sqr(b^2 - 4ac)} \ 2 and beta = {-b - sqr(b^2 - 4ac)}

Answer 2

SRY! alpha = {-b + sqr(b^2 - 4ac)} / 2a and beta = {-b - sqr(b^2 - 4ac)} / 2a

Answer 3

so product would be c/a and c=-21 and a=1 so it would be -21

Answer 4

Step 1: Find the two x-intercepts of the equation
Step 2: Multiply the x-intercepts together

X-intercepts can be found using the quadratic formula:
\[\frac{-b\pm\sqrt{b^2-4ac}}{2a}\]

Sub in your values:
\[\frac{-1*3\pm\sqrt{3^2-4*1*-21}}{2*1}\]
Find the two values. In this case, it was:
\(-6.32, 3.32\)
Multiply the numbers together:
\(-6.32*3.32=-20.9824\)

Answer 5

@Ahsome , no need to calculate b^2 - 4ac ;) look at my solution.

Answer 6

the product of the solutions of the equation ax^2 + bx + c = 0 is c/a

Answer 7

Any questions @babygirl65 ?

Answer 8

Do you know how to solve this equation because im not sure the one above is right @skullpatrol

Answer 9

@PFEH.1999

`\(\alpha=\frac{-b+\sqrt{b^2-4ac}}{2a}\)`


will give

\(\alpha=\frac{-b+\sqrt{b^2-4ac}}{2a}\)

Answer 10

Yes, I do. What does the question ask for @babygirl65 ?

Answer 11

what is the product of the 2 solutions of the equation x^2+3x-21=0 @skullpatrol

Answer 12

Yes the answer to that question can be seen by knowing that in the form x^2 + bx + c the number c is the product of the two solutions

Answer 13

thank you

Answer 14

Thanks for asking :-)

Answer 15

kudos to all the responses

Answer 16

what @PFEH.1999 said is great

for the form \(ax^2+bx+c=0\)

if we let \(\alpha\) and \(\beta\) be our two solutions we know the answers will be

\(\alpha=\frac{-b+\sqrt{b^2-4ac}}{2a}\) and
\(\beta=\frac{-b-\sqrt{b^2-4ac}}{2a}\)
then
\(\alpha*\beta=( \frac{-b-\sqrt{b^2-4ac}}{2a})*(\frac{-b+\sqrt{b^2-4ac}}{2a})= \frac{(-b)^2-b^2+4ac}{4a^2}=\frac{c}{a}\)

Answer 17

but @zzr0ck3r we don't need that form here :/

Answer 18

ha sorry was not trying to byte your style...

Answer 19

what do you mean you don't need that from here? this gives the general solution to any quadratics.

Answer 20

as long as the determinant is non negative...

Answer 21

All I'm saying is why not solve this question as a simpler special case of the general formula, for the sake of clarity @zzr0ck3r

Answer 22

you just asked a mathematician that question:P

Answer 23

:O

Answer 24

it was a very clever solution, I only typed it out because I thought the original poster was not great it latex, but I was wrong and waster my time.

Answer 25

Wow. That answer is GENIOUS. WOAH