System of two quadratics equation, P(x) and Q(x)If P(x)=ax2+bx+c and Q(x)=–ax2+dx+c,ac≠0, then the equation P(x). Q(x)=0 has(A) Exactly two real roots(B) At least two real roots(C) Exactly four real roots(D) No real roots

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System of two quadratics equation, P(x) and Q(x)If P(x)=ax2+bx+c and Q(x)=–ax2+dx+c,ac≠0, then the equation P(x). Q(x)=0 has(A) Exactly two real roots(B) At least two real roots(C) Exactly four real roots(D) No real roots

German Ferguson · Accepted Answer

Set PQ=(ax2+bx+c)(−ax2+dx+c) to zero, to get(ax2+bx+c)(−ax2+dx+c)=0.Check the discriminant for both P and Q.For P, you get DP=b2−4ac. For Q, you get DQ=d2+4ac.Suppose P has no real roots, such that DP=b2−4ac&amp;lt;0⇒4ac&amp;gt;b2. In that case, DQ=d2+4ac&amp;gt;0 and Q thus has two roots, so your polynomial PQ has at least two roots.Next, suppose P has 1 real root, in which case 4ac=b2 and so D1=d2+b2 which has 2 roots, so PQ has 3.Lastly, suppose P has 2 real roots, and thus b2&amp;gt;4ac, in which case Q may have zero, one, or two real roots.Therefore, PQ has at least two real roots.

Marquis Ibarra · Accepted Answer

You can do similar argue for your second equations also. And it will be real root&#039;s iff d2−4ac≥0.And now ac must be non zero . So either ac is positive or negative. So you can compare these two equation by considering both ac and you will find that discriminant remains always positive.

System of two quadratics equation, P(x) and Q(x) If

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