If the following equation has no real roots, what are the possible values of p...

Caleb Snyder

Caleb Snyder

Answered

2022-01-23

If the following equation has no real roots, what are the possible values of p for x2+(3p+1)xp=0?

Answer & Explanation

Palandriy0

Palandriy0

Expert

2022-01-24Added 14 answers

Step 1
We know that, the quadr. eqn,
ax2+bx+c=0 will have no real roots b24ac<0
Therefore, in our case, we must have,
(3p+1)24×1×(p)<0
9p2+6p+1+4p<0, i.e., 9p2+10p+1<0
(p+1)(9p+1)<0
 Case (1):(p+1)<0 and (9p+1)>0, or,
Case (2):(p+1)>0, and (9p+1)<0
Case (1):(p+1)<0 and (9p+1)>0
p<1 and p19. This is impossible.
Similarly, in Case (2), we have, p1, &, pM19, or
1<p<19
We conclude that, (1,19)

gekraamdbk

gekraamdbk

Expert

2022-01-25Added 13 answers

Step 1
If it has no real roots, then the discriminant of quadratic formula
Δ=(3p+1)2+4p<0
Δ=9p2+6p+1+4p=9p2+10p+1<0
Analyze this last expresion: represents a parabola for all values of p. Lets determine his roots
p=10±1003618=10±818=1 and 19
So in our case, the values of p for which the first equation has not real roots are in the interval (1, 19)
RizerMix

RizerMix

Expert

2022-01-27Added 437 answers

Step 1 If a quadratic equation has no real roots then its discriminant is less than zero. The discriminant of general quadratic equation ax2+bx+c=0 is b24ac We have the equation x2+(3p+1)xp=0 and its discriminant is (3p+1)24×1×(p)=9p2+6p+1+4p =9p2+10p+1=(9p+1)(p+1) Hence if x2+(3p+1)xp=0 has no real roots then (9p+1)(p+1)<0 Now there are two possiboloties 1) 9p+1<0 and p+1>0 which means p<19 and p>1S. This is possible when 1<p<19 2) 9p+1>0 and p+1<0 which means p>19 and p<1. But this is just not possible. Hence for x2+(3p+1)xp=0 to have no real roots we must have \(-1

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