Question

6.Given the quadratic equation mx^2-2mx-1=6x-m has two different roots,find the range of values of m. @ganeshie8 @SolomonZelman @UnkleRhaukus

Answer 1

I am not very sure how to do this, I would start of doing this....
(completing the square)

\(\large\color{slate}{ mx^2-2mx-1=6x-m }\)
\(\large\color{slate}{ mx^2-2mx-6x=1 -m }\)

\(\large\color{slate}{ mx^2-x(2m+6)=1 -m }\)

\(\large\color{slate}{ m(x^2-x\frac{2m+6}{m})=1 -m }\)

\(\large\color{slate}{ x^2-x\frac{2m+6}{m}=\frac{1 -m }{m}}\)

\(\large\color{slate}{ x^2-x\frac{2m+6}{m}+\frac{(m+3)^2}{m^2}=\frac{1 -m }{m}+\frac{(m+3)^2}{m^2}}\)

\(\large\color{slate}{ (x-\frac{m+3}{m})^2=\frac{1 -m }{m}+\frac{(m+3)^2}{m^2}}\)

\(\large\color{slate}{ (x-\frac{m+3}{m})^2=\frac{(1-m)^2+(m+3)^2}{m^2}}\)

\(\large\color{slate}{ (x-\frac{m+3}{m})^2=\frac{1-2m+m^2+m^2+6m+9}{m^2}}\)

\(\large\color{slate}{ (x-\frac{m+3}{m})^2=\frac{2m^2+4m+10}{m^2}}\)

Answer 2

answer is \[m>-\frac{ 9 }{ 7 }\]

Answer 3

why?

Answer 4

(not necessarily saying it is wrong, just want to see what you have done to obtain it)

Answer 5

no,i'm telling u the given answer.

Answer 6

Oh, idk how they did it.

Answer 7

(this is my gap in math... I would really like to see how someone else does this prob so that I know for future reference)

Answer 8

sorry that I couldn't help-:(

Answer 9

use b^2-4ac>0 @MARC_

Answer 10

before that

Answer 11

rearrange it in this form
ax^2+bx+c>0

Answer 12

A Quadratic Equation\[ax^2+bx+c=0\]
The Quadratic Formula\[x_{1,2}=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\]
The radicand of the quadratic formula is called
the Discriminant \(\Delta =b^2-4ac\)

if \(\Delta=0\) there is only one solution to the quadratic equation
if \(\Delta>0\) there are two solutions
if \(\Delta<0\) there are no real solutions


So, rearrange you equation of into the form
\[ax^2+bx+c=0\]
For two real roots the discriminate of this must greater then zero

Answer 13

\[mx^2-2mx-6x+m-1=0\]

Answer 14

@UnkleRhaukus

Answer 15

@ganeshie8

Answer 16

@ganeshie8

Answer 17

@hartnn

Answer 18

@TuringTest

Answer 19

@dan815

Answer 20

read what unklerhakus wrote

Answer 21

mx^2-2mx-1=6x-m
mx^2+(-2m-6)x-1+m=0
so under the sqrtb^2-4ac you have
b^2-4ac=(-2m-6)^2-4*m*(-1+m)
b^2-4ac>0 so that its not repeating or imaginary roots
(-2m-6)^2-4*m*(-1+m) > 0
solve for m

Answer 22

8m^2+28m+60>0

Answer 23

Is this correct so far ?
@dan815

Answer 24

@dan815

Answer 25

28m+36>0

Answer 26

m>-9/7

Answer 27

Got it finally !

Answer 28

http://www.wolframalpha.com/input/?i=%28-2m-6%29%5E2-4*m*%28-1%2Bm%29+%3E+0+

Answer 29

looks good!

Answer 30

Thank you @dan815 @UnkleRhaukus @ganeshie8 @eric_d @SolomonZelman