Question

The roots of the equation \[\left( {{p}^{4}}+{{q}^{4}} \right){{x}^{2}}+4pqrsx+{{r}^{4}}+{{s}^{4}}=0\] can be different, if real. True or False. True=1 False=0.

Answer 1

Hint: To solve the question, we have to apply the concept of value of discriminant of the quadratic equation for real roots greater than or equal to 0, to analyse the given quadratic equation. We have to rearrange the terms of the equation while solving to ease the procedure of solving.

Complete step-by-step answer:
The given quadratic equation is \[\left( {{p}^{4}}+{{q}^{4}} \right){{x}^{2}}+4pqrsx+{{r}^{4}}+{{s}^{4}}=0\]
We know that the roots of a quadratic equation are real when the discriminant of the equation is greater than or equal to 0.
The above statement mathematically is written as \[D\ge 0\], where D represent discriminant of the quadratic equation
The discriminant of the general quadratic equation \[a{{x}^{2}}+bx+c=0\] is given by \[D={{b}^{2}}-4ac\]
On comparing the given quadratic equation with the general quadratic equation, we get
\[a=\left( {{p}^{4}}+{{q}^{4}} \right),b=4pqrs,c={{r}^{4}}+{{s}^{4}}\]
By substituting the above values in discriminant of the quadratic equation formula, we get
\[\begin{align}
  & {{\left( 4pqrs \right)}^{2}}-4\left( {{p}^{4}}+{{q}^{4}} \right)\left( {{r}^{4}}+{{s}^{4}} \right) \\
 & =16{{p}^{2}}{{q}^{2}}{{r}^{2}}{{s}^{2}}-4\left( {{p}^{4}}{{r}^{4}}+{{q}^{4}}{{r}^{4}}+{{p}^{4}}{{s}^{4}}+{{q}^{4}}{{s}^{4}} \right) \\
 & =4\left( 4{{p}^{2}}{{q}^{2}}{{r}^{2}}{{s}^{2}}-{{p}^{4}}{{r}^{4}}-{{q}^{4}}{{r}^{4}}-{{p}^{4}}{{s}^{4}}-{{q}^{4}}{{s}^{4}} \right) \\
 & =-4\left( {{q}^{4}}{{r}^{4}}+{{p}^{4}}{{s}^{4}}-2{{p}^{2}}{{q}^{2}}{{r}^{2}}{{s}^{2}}+{{p}^{4}}{{r}^{4}}+{{q}^{4}}{{s}^{4}}-2{{p}^{2}}{{q}^{2}}{{r}^{2}}{{s}^{2}} \right) \\
\end{align}\]
We know the formula \[{{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab\]. Thus, by using this formula for simplifying the above equation, we get
\[=-4\left( {{\left( {{q}^{2}}{{r}^{2}}-{{p}^{2}}{{s}^{2}} \right)}^{2}}+{{\left( {{p}^{2}}{{r}^{2}}-{{q}^{2}}{{s}^{2}} \right)}^{2}} \right)\]
Thus, we get
\[-4\left( {{\left( {{q}^{2}}{{r}^{2}}-{{p}^{2}}{{s}^{2}} \right)}^{2}}+{{\left( {{p}^{2}}{{r}^{2}}-{{q}^{2}}{{s}^{2}} \right)}^{2}} \right)\ge 0\]
Since we know for real roots \[D\ge 0\]
\[4\left( {{\left( {{q}^{2}}{{r}^{2}}-{{p}^{2}}{{s}^{2}} \right)}^{2}}+{{\left( {{p}^{2}}{{r}^{2}}-{{q}^{2}}{{s}^{2}} \right)}^{2}} \right)\le 0\]
Since, the value of equation changes with change of sign of inequation.
The above inequation only stratifies when equal to 0, since the sum of squares of numbers is always positive.
Thus, we get \[4\left( {{\left( {{q}^{2}}{{r}^{2}}-{{p}^{2}}{{s}^{2}} \right)}^{2}}+{{\left( {{p}^{2}}{{r}^{2}}-{{q}^{2}}{{s}^{2}} \right)}^{2}} \right)=0\]
We know that the discriminant of quadratic is equal to 0 for equal roots. Thus, we can conclude that the roots of the given quadratic equation are equal.
Hence, the given statement is False = 0

Note: The possibility of mistake can be, not applying the concept of value of discriminant of the quadratic equation for real roots to analyse the given quadratic equation. The other possibility of mistake can be, not rearranging the terms of the equation while solving, which is an important step to ease the procedure of solving. The alternative way to solve the question is, after applying the discriminant condition for real roots for quadratic equation and solve further to find the relation between p, q, r, s, which will be useful to land at the answer.