Question
For what values of ‘p’ would the equation x2+2(p−1)x+p+5=0 possess at least one positive root?
Answer: Option A
:
A
For an equation to have positive roots D must be greater than '0'.
Now D from the equation =4(p−1)2−4(1)(p+5)=4p2−12p−16
=4(p2−3p−4)=4(p−4)(p+1)......(i)
x=[−b+√D]≥0
D≥b2
b=4(p−1)2
p2−3p−4≥p2+1−2p
p≤−5..........(2)
Taking intersection of equation 1 and 2
pE[−∞,−5]
So roots for eq. (i) are : -1 and 4 and comparing with ax2+bx+c , we have 'a' as>0 so function will be open ended in the upward direction.
So, , D is positive in region [−∞,−1] and [4,∞]. So for equation to posses at least one positive 'p' should either lie in [−∞,−1] or in [4,∞]. So answer is option (b).
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:
A
For an equation to have positive roots D must be greater than '0'.
Now D from the equation =4(p−1)2−4(1)(p+5)=4p2−12p−16
=4(p2−3p−4)=4(p−4)(p+1)......(i)
x=[−b+√D]≥0
D≥b2
b=4(p−1)2
p2−3p−4≥p2+1−2p
p≤−5..........(2)
Taking intersection of equation 1 and 2
pE[−∞,−5]
So roots for eq. (i) are : -1 and 4 and comparing with ax2+bx+c , we have 'a' as>0 so function will be open ended in the upward direction.
So, , D is positive in region [−∞,−1] and [4,∞]. So for equation to posses at least one positive 'p' should either lie in [−∞,−1] or in [4,∞]. So answer is option (b).
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