Question

Find the number of integral value of `m` for which exactly one root of the equation `x^2-2mx+m^2-1 = 0` lies in the interval `(-2,4)`?

A) 6
B) 5
C) 4
D) None

Answer 1

The answer is (C) `4`.

Explanation:

The quadratic equation given to you looks like this

`x^2 - 2mx + m^2 - 1 = 0`

Use the quadratic formula to find the two solutions for `x`

`x_(1,2) = (- (-2m) +- sqrt( (-2m)^2 - 4 * 1 * (m^2-1)))/(2 * 1)`

`x_(1,2) = (2m +- sqrt(color(red)(cancel(color(black)(4m^2))) - color(red)(cancel(color(black)(4m^2))) + 4))/2`

`x_(1,2) = (2m +- sqrt(4))/2 implies {(x_1 = (2m + 2)/2 = m + 1), (x_2 = (2m - 2)/2 = m-1) :}`

Now, you know that you must find the number of integer values of `m` for which exactly one root of this quadratic equation lies in `(-2,4)`.

In other words, you're looking for the number of values of `m` that will make

`-2 < m + 1 < 4" " ul(and) " " 4 >= m -1 >= -2`

You will have

`-3 < m < 3 " " ul(and) " " 5 >= m >= -1`

For `m in ZZ`, the first compound inequality has `5` solutions

`m in {-2, -1, 0 ,1 ,2}`

Similarly, the second compound inequality has `7` solutions--keep in mind that the second compound inequality has greater than or equal to signs!

`m in {-1, 0, 1, 2, 3, 4, 5}`

In order to have

`m in {-2, -1, 0 ,1 ,2} " " ul(and)" "{-1, 0, 1, 2, 3, 4, 5}`

you need

`m in {-1, 0, 1, 2}`

Therefore, a total of `4` integer values for `m` will make exactly one root lie in `(-2,4)`.

Notice that you will get the same result, i.e. `4` integer values for `m`, if you take

`-2 < m-1<4" "ul(and)" "4 >= m+1 >= -2`