# 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)#?

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A) 6

B) 5

C) 4

D) None

A) 6

B) 5

C) 4

D) None

##### 1 Answer

The answer is **(C)**

#### 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_(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 **one root** of this quadratic equation lies in

In other words, you're looking for the number of values of

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

You will have

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

For **solutions**

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

Similarly, the second compound inequality has **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 **integer values** for

Notice that you will get the same result, i.e. **integer values** for

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