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

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Palandriy0 · Accepted Answer

Step 1 We know that, the quadr. eqn, ax2+bx+c=0 will have no real roots ⇔b2−4ac&amp;lt;0 Therefore, in our case, we must have, (3p+1)2−4×1×(−p)&amp;lt;0 ∴9p2+6p+1+4p&amp;lt;0, i.e., 9p2+10p+1&amp;lt;0 ∴(p+1)(9p+1)&amp;lt;0 ∴ Case (1):(p+1)&amp;lt;0 and (9p+1)&amp;gt;0, or, Case (2):(p+1)&amp;gt;0, and (9p+1)&amp;lt;0 Case (1):(p+1)&amp;lt;0 and (9p+1)&amp;gt;0 ∴p&amp;lt;−1 and p≻19. This is impossible. Similarly, in Case (2), we have, p≻1, &amp;amp;, pM−19, or −1&amp;lt;p&amp;lt;−19 We conclude that, ∈(−1,−19)

gekraamdbk · Accepted Answer

Step 1  If it has no real roots, then the discriminant of quadratic formula  Δ=(3p+1)2+4p&amp;lt;0  Δ=9p2+6p+1+4p=9p2+10p+1&amp;lt;0  Analyze this last expresion: represents a parabola for all values of p. Lets determine his roots  p=−10±100−3618=−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 · Accepted Answer

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 b2−4ac We have the equation x2+(3p+1)x−p=0 and its discriminant is (3p+1)2−4×1×(−p)=9p2+6p+1+4p =9p2+10p+1=(9p+1)(p+1) Hence if x2+(3p+1)x−p=0 has no real roots then (9p+1)(p+1)&amp;lt;0 Now there are two possiboloties 1) 9p+1&amp;lt;0 and p+1&amp;gt;0 which means p&amp;lt;−19 and p&amp;gt;−1S. This is possible when −1&amp;lt;p&amp;lt;−19 2) 9p+1&amp;gt;0 and p+1&amp;lt;0 which means p&amp;gt;−19 and p&amp;lt;−1. But this is just not possible. Hence for x2+(3p+1)x−p=0 to have no real roots we must have (-1

If the following equation has no real roots, what are

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