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

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


2 Answers

25763 Points
one year ago
  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=1 or x=-1 ====> It satisfies the given range

now substitute m=1, Then



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



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

Aditya Gupta
2075 Points
one year ago
note that aruns ans is absolutely wrong and also assumes that m is an integer. he has probably copied the ans from somewhere.
correct method:
given x^2 – 2mx + m^2= 1
or (x – m)^2= 1
or x= m ± 1
so, roots are m+1 and m – 1.
now, as m lies in ( – 1, 3), so m – 1 lies in ( – 2, 2) and m+1 lies in (0, 4).
so overall both the roots lie in the union of ( – 2, 2) and (0, 4).
so, the roots of x^2-2mx+m^2-1=0 lies in (-2,4).
kindly approve :)

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