Question

can someone work this out and see if my answer is correct? Discriminant

find the value of c where the quadratic equation has an equal root.(= 0 )

3cx^2 + 4(c+1)x + (c+1)=0

my answer is c=-1 but i think it not right i think

Answer 1

This is impossible to solve, because there are 2 variables.

Answer 2

Its not to solve x... It to find it root and the root is 0 and then solve it for c.. but i think i did something wrong when i solved it.

But i know where your coming from but it not that quadratic solving

Answer 3

If x = 0, than you are correct, and c = -1.

Answer 4

no.. please understand me saying 0 im talking about the "root" is equal to zero. To solve this problem it's to use discriminant of a quadratic. This is part of Discriminant.

Answer 5

root is not x .. root is to predict whether there will be two,one or no solution to my equation

Answer 6

so you're saying that you're finding c to see how many solutions there are in the equation?

Answer 7

my process of factoring out this problem so far

Answer 8

no as stated in my question it has an equal root meaning there will be one solution, where b^2-4ac=0. and that's a good progress your doing but it in a different way of solving. to solve this it is used by b^2-4ac=0 and swap in the value from my equation. so after the swapping i then go lost.

Answer 9

your image(work) is good but your putting in plus instead of -.. except if i am totally wrong about this

Answer 10

well your given problem, you put it all in plus.

Answer 11

so your solving it like b^2+4ac not b^2-4ac

Answer 12

grouping factors**

Answer 13

i did that at first but my teacher said not to

Answer 14

Working through it I get the discriminant = \[-2c ^{2}+5c+1\]
Using the quadratic equation on that, I get \[c=-(\frac{ 5 }{ 4 }\pm \sqrt{17})\]

Answer 15

can you show me how you work this out?

Answer 16

sorry for asking too much but i just wanna know whered u ge the −2c^2+5c+1

Answer 17

I made a mistake, but here are correct steps.
\[3cx^{2c}+(4c+4)x+(+1)=0\](using quadratic equation)\[x=\frac{ -(4c+4) \pm \sqrt{(4c+4)^{2}-4*3c*(c+1)} }{ 6c }\](taking determinant = 0)\[(4c+4)^{2}-(12c^2+12c)=0\]\[16c^{2}-12c^{2}+32c+12c+16=0\]\[4c^{2}+44c+16=0\]\[c^2+11c+4=0\](using quadratic equation again)\[c=\frac{ -11\pm \sqrt{11^{2}-4*1*4} }{ 2 }\]\[c=\frac{ -11\pm \sqrt{121-16} }{ 2 }\]\[c=\frac{ -11\pm \sqrt{105} }{ 2 }\]

Answer 18

Good night.

Answer 19

i see.. theres this little mistake you made around your fisrt step where b^2-4ac and the a you squared the c but the x is the only variable which is squared not c.

Answer 20

a = 3c, b = (4c+4), c = c+1
b^2-4ac = (4c+4)^2 - 4*3c*(c+1)
I don't see what I did wrong, except I put a v+ instead of a - in my 4th step.

Answer 21

A +, not a v+.

Answer 22

@puzzler7 your expansion here is not correct
\[(4c+4)^{2}-(12c^2+12c)=0\\16c^2+32c+16 -12c^2-12c=0\]
you made mistake at the last term. It's -12c not 12c as you did

Answer 23

I know, I just noted that.

Answer 24

so that, the discriminant is
\[4c^2+20c+16=0\] which gives us 2 roots: c = -1 and c=-4

Answer 25

thank you oops.. i guess my -1 ans is correct. thank you very much..

Answer 26

:)

Answer 27

puzzler7 thank you to u too :D

Answer 28

Welcome. Sorry for all the mistakes.

Answer 29

oy,, dont say that, I was the one who didn't know anything and you helped a lot. Thank you :D