Question

Show‌ ‌that‌ ‌if‌ ‌p,‌ ‌q,‌ ‌r‌ ‌and‌ ‌s‌ ‌are‌ ‌real‌ ‌numbers‌ ‌and‌ ‌$pr=2\left(‌ ‌q+s‌ ‌\right)$‌ ‌then‌ ‌at‌ ‌least‌ ‌one‌ ‌of‌ ‌the‌ ‌ equations‌ ‌${{x}^{2}}+px+q=0,{{x}^{2}}+rx+s=0$‌ ‌has‌ ‌real‌ ‌roots?‌ ‌

Answer 1

Hint: First of all, we will find the discriminant of the two quadratic equations given in the above problem. We know that the formula for the discriminant of the quadratic equation for the quadratic equation say $a{{x}^{2}}+bx+c=0$ is equal to ${{b}^{2}}-4ac$. Then add the two discriminants and then deploy the given equation $pr=2\left( q+s \right)$ into this addition of these discriminants. Hence, from this addition we can show that at least one of the given two quadratic equations possess real roots.

Complete step-by-step solution:
The given quadratic equations are as follows:
$\begin{align}
  & {{x}^{2}}+px+q=0, \\
 & {{x}^{2}}+rx+s=0 \\
\end{align}$
We are going to find the discriminant of the above two equations. Let us take a general quadratic equation of the following form:
$a{{x}^{2}}+bx+c=0$
Now, discriminant for the above quadratic equation is as follows:
${{b}^{2}}-4ac$
Using the above formula for discriminant in the above two equations we get,
Let us say that discriminant of the first and the second equations are ${{D}_{1}}\And {{D}_{2}}$ respectively so writing the discriminant for the two equations:
$\begin{align}
  & {{D}_{1}}={{p}^{2}}-4q; \\
 & {{D}_{2}}={{r}^{2}}-4s \\
\end{align}$
Adding the above two equations we get,
$\begin{align}
  & \Rightarrow {{D}_{1}}+{{D}_{2}}={{p}^{2}}-4q+{{r}^{2}}-4s \\
 & \Rightarrow {{D}_{1}}+{{D}_{2}}={{p}^{2}}+{{r}^{2}}-4s-4q \\
\end{align}$
Adding and subtracting 2pr in the above equation we get,
$\Rightarrow {{D}_{1}}+{{D}_{2}}={{p}^{2}}+{{r}^{2}}-2pr+2pr-4s-4q$
Now, we can write ${{p}^{2}}+{{r}^{2}}-2pr={{\left( p-r \right)}^{2}}$ in the above equation and we get,
$\Rightarrow {{D}_{1}}+{{D}_{2}}={{\left( p-r \right)}^{2}}-2\left( pr-2\left( s+q \right) \right)$
In the above problem, it is given that $pr=2\left( q+s \right)$ so $pr-\left( 2s+2q \right)=0$, using this relation in the above equation we get,
$\begin{align}
  & \Rightarrow {{D}_{1}}+{{D}_{2}}={{\left( p-r \right)}^{2}}-0 \\
 & \Rightarrow {{D}_{1}}+{{D}_{2}}={{\left( p-r \right)}^{2}} \\
\end{align}$
Now, from the above equation you can see that addition of the two discriminants is positive this means at least one of the two discriminants must be positive. And we know that, for a quadratic equation to have real roots, its discriminant must be greater than or equal to 0.
Hence, we have shown that at least one of the two quadratic equations must possess real roots when p, q, r and s are real numbers and $pr=2\left( q+s \right)$.

Note: To solve the above problem, you need to know about what is discriminant and the formula for the discriminant. Also, we should know using the value of discriminant, how we come to know if a quadratic equation possesses real roots or not. If you don’t know these properties then you cannot move further in this problem.