All the values of m for which both roots of the equation x^2 − 2mx + m

x^2−2mx+m^2=1 -
(x-m)^2=1
x-m=±1
x=m±1

Roots (x1, x2) are between -2 and 4
-2<x<4
-2<m±1<4
-2∓1<m<4∓1
==> -3<m<3 or -1<m<5.
Superposing both ranges, one gets -1<m<3

FINAL ANSWER IS (C) −1<m<3

All the values of m for which both roots of the equation x^2−2mx+m^2−1=0 are greater than - 2 but less than 4, lie in the interval
(x-m)^2=1
x-m = 1 or x-m=-1
x = m + 1 or x = m-1
m-1>-2
m>-1
m+1<4
m<3
-1<m<3

A. m<−2

B. −2<m<0

C. −1<m<3

D. 1<m<4

E. m>4

IMO C

The roots of the equation ax2 + bx + c = 0 are given by formula:
α = (-b-√b2-4ac)/2a
β = (-b+√b2-4ac)/2a
Where β and α Are roots of the equation
Hence here solving we get 2 roots as
m+1 and m-1
Both these lie in between -2 and 4
=> -2<m+1<4 & -2<m-1<4
=> -3<m<3 & -1<m<5
Hence solving above inequality
M should lie in between -1 & 3
Hence, I think answer should be C
-1<m<3

Posted from my mobile device

(x-m)^2-1=0
(x-m)^2-1^2=0
(x-m-1)(x-m+1)=0
x=m+1, x=m-1
-2<m-1<m+1<4
-1<m<3

Ans (C)

for the given quadratic ; the value of the product of roots = 1 ; sum of roots= 2m & m^2
given that value of roots is -2<m<4
only option B ; −2<m<0 suffices the given range of roots
IMO B

All the values of m for which both roots of the equation x2−2mx+m2−1=0 are greater than - 2 but less than 4, lie in the interval

A. m<−2m<−2

B. −2<m<0−2<m<0

C. −1<m<3−1<m<3

D. 1<m<41<m<4

E. m>4

(x - m)^2 - 1 = 0
(x - m +1) (x - m -1)= 0
x = m - 1 or x = m+1
-2 < m-1 < 4
-1 < m < 5
or
-2 < m+ 1 < 4
-3 < m < 3
So, m covers the range starting from -1 to 3. C is the answer.

All the values of m for which both roots of the equation \(x^2−2mx+m^2−1=0\) are greater than - 2 but less than 4, lie in the interval

A. m < −2

B. −2 < m < 0

C. −1 < m < 3

D. 1 < m < 4

E. m > 4

For quadratic equation \(ax^2 + bx + c = 0\) x = \(\frac{-b ± \sqrt{b^2 - 4ac}}{2a}\)
Let \(x^2−2mx+m^2−1=0\) has roots A and B.
Then
x = \(\frac{-(-2m) ± \sqrt{(-2m)^2 - 4*1*(m^2-1)}}{2*1}\)

x = \(m ± \sqrt{2m^2 - 1}\)

As per question -2 < x < 4, So
-2 < \(m - \sqrt{2m^2 - 1}\) < 4 OR -2 < \(m + \sqrt{2m^2 - 1}\) < 4

Option A and E are out since entity involved is square which would make the value of \(m - \sqrt{2m^2 - 1}\) less than -2 OR greater than 4.

Since B and C share a common inequality -1 < m < 0, lets check for m = \(\frac{-1}{2}\)
\((-\frac{1}{2}) - \sqrt{2(-\frac{1}{2})^2 - 1}\) OR \((-\frac{1}{2}) - \sqrt{2(-\frac{1}{2})^2 - 1}\)
which gives imaginary roots thus its not applicable solution.

So, Answer is D.

Hi! I understand that if x=m+1, then -3<m<3 and if x=m-1, then -1<m<5. But how come the answer is -1<m<3? Since the question asks for the interval in which ALL values of m are in, shouldn't it be -3<m<5 as that covers the entire range of m? Could someone please explain why C is correct? Thanks in advance!

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