Ifm belongs to (-1,3) then prove that the roots of x^2-2mx+m^2-1=0 lies in (-2,4)

1. d> 0: 2 distinct real roots: factors over the reals
2. d
3. d = 0: 1 real root with 2: factors over the reals as a square.

Here d=b^2–4ac as given equation is having two roots we follow above rules , First we need to know the basics before solving this question

Given range -1m values can be (0,1,2)

Given Equation x^2–2mx-1=0

We need to prove that x values lies in between (-2,4)=====> This means x values can be (-1,0,1,2,3) and they can be complex roots also

x^2–2mx-1=0….substitute m=0 then

x^2–2(0)(x)-1=0

x^2=1

x=1 or x=-1 ====> It satisfies the given range

now substitute m=1, Then

x^2–2(1)x-1=0

x^2–2x-1=0

to find roots use this formulae -b+-suart(b^2–4ac)/2*a=====>

2+squart(8)/2 and 2-squart(8)/2 ==> 2(1+squart(2))/2 and 2(1-squart(2))/2

On simpler calculation we get roots as 1+squrt(2),1-squrt(2) This lies in the above mentioned range of x(-1,0,1,2,3)

Now substitute m=2,Then

x^2–2(2)x-1=0

x^2–4x-1=0

Again to find roots use the same formulae -b+-suart(b^2–4ac)/2*a

2+squart(20)/2 and 2-squart(20)/2 ====> 2(1+squart(5))/2 and 2(1-squart(5))/2

so roots are 1+squart(5) and 1-squart(5) ===> This satisfies the desired range but one of the root is not satisfied as the value is just above 3

Note : Here to find roots i used formaule -b+-squart(b^2–4ac)/2*a

where b^2–4ac=discriminant