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↧↧↧ Detailed Video Solution to the Problem ↧↧↧



We need to find how many real solutions exist for the equation \(x^2 – 11|x| - 60 = 0\)

As we have |x| in the equation so we will have two cases

As we have |x| in the equation so we will have two cases
-Case 1: x ≥ 0
=> |x| = x
=> \(x^2 – 11x - 60 = 0\)
=> \(x^2 + 4x - 15x - 60 = 0\)
=> x*(x + 4) -15*(x + 4) = 0
=> (x + 4) * (x - 15) = 0
=> x = -4, 15

But condition was x ≥ 0
=> x = 15 is a SOLUTION
-Case 2: x < 0
=> |x| = -x
=> \(x^2 + 11x - 60 = 0\)
=> \(x^2 + 15x - 4x - 60 = 0\)
=> x*(x + 15) -4*(x + 15) = 0
=> (x + 15) * (x - 4) = 0
=> x = 4, -15

But condition was x < 0
=> x = -15 is a SOLUTION

So, there are 2 real solutions of above equation. (x = -15, 15)

So, Answer will be C
Hope it helps!

Watch the following video to MASTER Absolute Values

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Take the discriminant of the equation

b^2-4(a)(c) = (+/- 11)^2 - (4)(1)(-6) > 0

Determinant is positive, therefore two real roots are possible from the equation.
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assume mod(x)=y
y^2-11y-60=0
(y-15)(y+4)=0
y=15 or y=-4
since mod(x) can never be -ve so discarding y=-4
so mod(x)=15
x==15 or -15
hence two real solutions
hence C
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