If a, b, c are real numbers such that ac ≠ 0,

Solution:

The given equations are

$a x^{2}+b x+c=0$........(1)

 

$-a x^{2}+b x+c=0$......(2)

Roots are simultaneously real

Let $D_{1}$ and $D_{2}$ be the discriminants of equation (1) and (2) respectively,

 

Then,

$D_{1}=(b)^{2}-4 a c$

$=b^{2}-4 a c$

And

$D_{2}=(b)^{2}-4 \times(-a) \times c$

$=b^{2}+4 a c$

Both the given equation will have real roots, if $D_{1} \geq 0$ and $D_{2} \geq 0$.

Thus,

$b^{2}-4 a c \geq 0$

$b^{2} \geq 4 a c$ ......(3)

And,             

$b^{2}+4 a c \geq 0$ .......(4)

Now given that are real number and as well as from equations (3) and (4) we get

At least one of the given equation has real roots

Hence, proved