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

If $$P(x) = ax^2 + bx + c$$ and $$Q(x) = – ax^2 + dx + c$$, $$ac \ne 0$$, 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

My approach is as follow Let $$T(x)=P(x).Q(x)$$

$$T\left( x \right) = - a^2{x^4} + a\left( {d - b} \right){x^3} + \left( {bd} \right){x^2} + c\left( {d + b} \right)x + {c^2}$$

Not able to approach from here

If the coefficient of $$x^2$$ and the constant term of a quadratic equation are of opposite sign $$($$both non-zero$$)$$ then discriminant of the quadratic equation is positive, which means roots are real.

So either $$P(x)$$ or $$Q(x)$$ has coefficient of $$x^2$$ and the constant term with opposite sign $$($$both are non-zero$$)$$. So one of them has $$2$$ real roots.

Therefore $$P(x)Q(x)=0$$ has at least $$2$$ real roots.

We can also solve using your approach
$$T\left( x \right) = - a^2{x^4} + a\left( {d - b} \right){x^3} + \left( {bd} \right){x^2} + c\left( {d + b} \right)x + {c^2}$$
$$T(0)>0$$ and for sufficiently large $$k$$( $$k$$ is positive) we have $$T(k)<0$$ and $$T(-k)<0$$.
So there should be a real root in the interval $$(-k,0)$$ and $$(0,k)$$ so at least $$2$$ real roots exist.

Set $$PQ=\left( ax^2+bx+c\right) \left( -ax^2+dx+c\right)$$ to zero, to get $$\left( ax^2+bx+c\right) \left( -ax^2+dx+c\right)=0.$$

Check the discriminant for both $$P$$ and $$Q$$.

For $$P$$, you get $$D_P=b^2-4ac$$. For $$Q$$, you get $$D_Q=d^2+4ac$$.

Suppose $$P$$ has no real roots, such that $$D_P=b^2-4ac < 0 \implies4ac>b^2$$. In that case, $$D_Q=d^2+4ac>0$$ 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 $$4ac=b^2$$ and so $$D_1 = d^2+b^2$$ which has 2 roots, so $$PQ$$ has 3.

Lastly, suppose $$P$$ has 2 real roots, and thus $$b^2>4ac$$, in which case $$Q$$ may have zero, one, or two real roots.

Therefore, $$PQ$$ has at least two real roots.

it must option b. Why ? Let's see .

Your first $$ax^2 + bx +c$$ has real root's when the value of discriminant attached to it positive . So $$b^2 - 4ac >_ 0$$.

You can do similar argue for your second equations also. And it will be real root's iff $$d^2 - 4ac ≥ 0$$

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.

The parabolas represented by the quadratic functions $$\ P(x) \ = \ ax^2 + bx + c \$$ and $$\ Q(x) \ = \ – ax^2 + dx + c \ \ , \ \ ac \ne 0 \ \ ,$$ "open" in opposite "vertical" directions and have a common $$\ y-$$intercept. The absolutely minimal number of $$\ x-$$intercepts that could be arranged is one, by placing the vertices of both parabolas at the origin. But these are $$\ y \ = \ ax^2 \$$ and $$\ y \ = \ -ax^2 \ \ ,$$ which are not permitted by the condition $$\ ac \ne 0 \ \ ; \$$ a "vertical shift" in either direction necessarily produces two $$\ x-$$intercepts, and so two real zeroes for $$\ (ax^2 + c)·(-ax^2 + c) \ \ .$$

Thus, the unavoidable minimum number of real zeroes for $$\ P(x)·Q(x) \$$ is two. We can arrange for more than two by placing the two parabolas so that they are tangent at the origin at a point other than their vertices, with $$\ y \ = \ ax^2 + bx \$$ and $$\ y \ = \ -ax^2 + bx \ \ ( d = b ) \ \ .$$ With their vertices then at $$\ x \ = \ \pm \frac{b}{2a} \ \ ,$$ there are three $$\ x-$$intercepts, including one at the origin. Again, we may not have $$\ c = 0 \ \ ,$$ but a "vertical shift" in either direction, that is kept small enough that the vertices remain on opposite sides of the $$\ x-$$axis, then leads to four $$\ x-$$intercepts (the one at the origin "bifurcating" into two).

We demonstrate geometrically then that we must have at least two zeroes for $$\ P(x)·Q(x) \$$ and may easily obtain up to four [choice $$\ \mathbf{(B)}$$ ] . If we generally "break" the symmetry of the vertices about the origin described in the previous paragraph, we can easily obtain more than two $$\ x-$$intercepts for the parabolas.

We can also see this by putting the polynomials into "vertex form": $$a·\left(x \ + \ \frac{b}{2a} \right)^2 \ + \ \left(c \ - \ \frac{b^2}{4a} \right) \ \ , \ \ -a·\left(x \ - \ \frac{d}{2a} \right)^2 \ + \ \left(c \ + \ \frac{d^2}{4a} \right)$$ and finding their zeroes from $$\left(x \ + \ \frac{b}{2a} \right)^2 \ \ = \ \ -\frac{1}{a}·\left(c \ - \ \frac{b^2}{4a} \right) \ \ , \ \ \left(x \ - \ \frac{d}{2a} \right)^2 \ \ = \ \ \frac{1}{a}·\left(c \ + \ \frac{d^2}{4a} \right) \ \ .$$ Often, it will be the case that the right side of one of these equations is positive while the right side of the other is negative, giving us only two real zeroes. But it is also possible for both right sides to be positive when $$\ \frac{b^2}{4a^2} \ > \ \frac{c}{a} \ > \ -\frac{d^2}{4a^2} \ \ ,$$ producing four real zeroes. [This is equivalent to the condition for the polynomial discriminants.]