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04 Jan 2022, 04:15
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C
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Difficulty:
65%
(hard)
Question Stats:
56% (01:50) correct
44%
(01:50)
wrong
based on 712
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How many real solutions exist for the equation \(x^2 – 11|x| - 60 = 0\)?
A. 4
B. 3
C. 2
D. 1
E. 0
A. 4
B. 3
C. 2
D. 1
E. 0
05 Jan 2022, 22:33
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\(x^2–11|x|−60=0\)
Case 1: \(x ≥ 0\)
\(x^2–11x−60=0\)
When ever the sign of the constant C in the quadratic equation of general form ax^2 + bx + c =0 is negative, then two roots will be of opposite sign,i.e one root is +ve and other one is -ve.
Since the condition of case 1 is x ≥ 0 , we will consider only the positive root.
Case 2: \(x< 0\)
\(x^2+11x−60=0\)
Same concept we can apply in case 2 as well. The sign of constant C is -ve in the quadratic equation, so here also we have 2 roots , one's + ve and the other -ve.
Since x should be less than zero, we will consider only the -ve root.
Combining case 1 and 2, Only two real solutions exist for the equation.
Option C is the answer.
Thanks,
Clifin J Francis
Case 1: \(x ≥ 0\)
\(x^2–11x−60=0\)
When ever the sign of the constant C in the quadratic equation of general form ax^2 + bx + c =0 is negative, then two roots will be of opposite sign,i.e one root is +ve and other one is -ve.
Since the condition of case 1 is x ≥ 0 , we will consider only the positive root.
Case 2: \(x< 0\)
\(x^2+11x−60=0\)
Same concept we can apply in case 2 as well. The sign of constant C is -ve in the quadratic equation, so here also we have 2 roots , one's + ve and the other -ve.
Since x should be less than zero, we will consider only the -ve root.
Combining case 1 and 2, Only two real solutions exist for the equation.
Option C is the answer.
Thanks,
Clifin J Francis
General Discussion
04 Jan 2022, 06:55
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|x|=x for x>0
-x.for x<0
Case I (for x>0)
x²-11x-60=0
X=15
Case II (for x<0)
x²+11x-60=0
X=-15
So only 2 real values are possible for the above equation..
C is the answer.
Posted from my mobile device
-x.for x<0
Case I (for x>0)
x²-11x-60=0
X=15
Case II (for x<0)
x²+11x-60=0
X=-15
So only 2 real values are possible for the above equation..
C is the answer.
Posted from my mobile device
