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If both roots of the equation x2 - 2ax + a2 -1 = 0 lie in the interval (-3,4) then sum of the integral parts of ‘a’ is
- a)0
- b)2
- c)4
- d)–1
Correct answer is option 'A'. Can you explain this answer?
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Verified Answer
If both roots of the equation x2 - 2ax + a2 -1 = 0 lie in the interval...
x2 - 2ax + a2 -1 = 0 ⇒ (x - a)2 = 1
Roots are a - 1 and a + 1
Both roots lie in (-3, 4) ∴ a - 1 > -3, a + 1 < 4
Ie -2 < a < 3 ⇒ [a] = -2, -1, 0, 1, 2 whose sum is zero.
Roots are a - 1 and a + 1
Both roots lie in (-3, 4) ∴ a - 1 > -3, a + 1 < 4
Ie -2 < a < 3 ⇒ [a] = -2, -1, 0, 1, 2 whose sum is zero.
Most Upvoted Answer
If both roots of the equation x2 - 2ax + a2 -1 = 0 lie in the interval...
The possible values of 'a' is 5.
We can start by finding the discriminant of the given quadratic equation:
b^2 - 4ac = (-2a)^2 - 4(1)(a^2 - 1) = 4a^2 - 4a^2 + 4 = 4
Since the discriminant is positive, the quadratic equation has two real roots. Let's call them x1 and x2:
x1 = a + sqrt(a^2 - 1)
x2 = a - sqrt(a^2 - 1)
For both roots to lie in the interval (-3,4), we need the following conditions to be true:
-4 < a="" +="" sqrt(a^2="" -="" 1)="" />< 3="" (since="" the="" larger="" root="" cannot="" exceed="" />
-4 < a="" -="" sqrt(a^2="" -="" 1)="" />< 3="" (since="" the="" smaller="" root="" cannot="" be="" less="" than="" />
Simplifying these inequalities, we get:
-3 < sqrt(a^2="" -="" 1)="" />< 4="" -="" />
-3 < sqrt(a^2="" -="" 1)="" />< a="" +="" />
Squaring both sides of these inequalities (remembering to consider both positive and negative roots), we get:
1 < a^2="" />< (a="" -="" 1)(a="" -="" />
1 < a^2="" />< (a="" +="" 1)(a="" -="" />
The solutions to these inequalities are:
-2 < a="" />< -1="" or="" a="" /> 8
-1 < a="" />< 0="" or="" a="" /> 4
Taking the intersection of these solution sets, we get:
-1 < a="" />< 0="" or="" a="" /> 8
The possible values of 'a' that satisfy this condition and have integral parts are:
a = -1, 0, 9
The sum of the integral parts of these values is 5.
We can start by finding the discriminant of the given quadratic equation:
b^2 - 4ac = (-2a)^2 - 4(1)(a^2 - 1) = 4a^2 - 4a^2 + 4 = 4
Since the discriminant is positive, the quadratic equation has two real roots. Let's call them x1 and x2:
x1 = a + sqrt(a^2 - 1)
x2 = a - sqrt(a^2 - 1)
For both roots to lie in the interval (-3,4), we need the following conditions to be true:
-4 < a="" +="" sqrt(a^2="" -="" 1)="" />< 3="" (since="" the="" larger="" root="" cannot="" exceed="" />
-4 < a="" -="" sqrt(a^2="" -="" 1)="" />< 3="" (since="" the="" smaller="" root="" cannot="" be="" less="" than="" />
Simplifying these inequalities, we get:
-3 < sqrt(a^2="" -="" 1)="" />< 4="" -="" />
-3 < sqrt(a^2="" -="" 1)="" />< a="" +="" />
Squaring both sides of these inequalities (remembering to consider both positive and negative roots), we get:
1 < a^2="" />< (a="" -="" 1)(a="" -="" />
1 < a^2="" />< (a="" +="" 1)(a="" -="" />
The solutions to these inequalities are:
-2 < a="" />< -1="" or="" a="" /> 8
-1 < a="" />< 0="" or="" a="" /> 4
Taking the intersection of these solution sets, we get:
-1 < a="" />< 0="" or="" a="" /> 8
The possible values of 'a' that satisfy this condition and have integral parts are:
a = -1, 0, 9
The sum of the integral parts of these values is 5.
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If both roots of the equation x2 - 2ax + a2 -1 = 0 lie in the interval (-3,4) then sum of the integral parts of ‘a’ isa)0b)2c)4d)–1Correct answer is option 'A'. Can you explain this answer?
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