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If the equation, kx^2+4x+(5-k)=0
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21 Jul 2019, 05:54
21 Jul 2019, 05:54
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58% (01:31) correct
41% (02:12) wrong based on 31 sessions
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If the equation, \(kx^2+4x+(5-k)=0\) has real roots, find the inequality followed by k?
A) \(k^2-5k+4>0\)
B) \(k^2-4k+5>0\)
C) \(k^2-5k+4<0\)
D) \(k^2-4k+5<0\)
E) \(k^2-4k+4<0\)
A) \(k^2-5k+4>0\)
B) \(k^2-4k+5>0\)
C) \(k^2-5k+4<0\)
D) \(k^2-4k+5<0\)
E) \(k^2-4k+4<0\)
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Joined: 20 Jun 2019
Posts: 181
Given Kudos: 41
GRE 1: Q160 V161
GRE 2: Q165 V159
Re: If the equation, kx^2+4x+(5-k)=0
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10 Mar 2020, 23:29
10 Mar 2020, 23:29
Can someone please explain this?
Re: If the equation, kx^2+4x+(5-k)=0
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11 Mar 2020, 23:30
11 Mar 2020, 23:30
2
equation, kx2+4x+(5−k)=0
Remember the quadratic formula.
( -b +/- sqrt(b^2-4ac) ) / 2a = the roots
( -4 +/- sqrt(16-4k(5-k) ) / 2a =
( -4 +/- sqrt(16-20k+4k^2) / 2a =
Note that we only care about making sure
That the sqrt is positive so I’m only going to look at
That
16-20k+4k^2 > 0 when?
4(4-5k+k^2) > 0
4-5k+k^2 > 0
Which is the same as A.
Posted from my mobile device
Remember the quadratic formula.
( -b +/- sqrt(b^2-4ac) ) / 2a = the roots
( -4 +/- sqrt(16-4k(5-k) ) / 2a =
( -4 +/- sqrt(16-20k+4k^2) / 2a =
Note that we only care about making sure
That the sqrt is positive so I’m only going to look at
That
16-20k+4k^2 > 0 when?
4(4-5k+k^2) > 0
4-5k+k^2 > 0
Which is the same as A.
Posted from my mobile device
