Question

If $p,q,r\in R$ and the quadratic equation $p{{x}^{2}}+qx+r=0$ has no real root, then
1. $p(p+q+r)>0$
2. $r(p+q+r)>0$
3. $q(p+q+r)>0$
4. $(p+q+r)>0$

Answer 1

In question it is given that $p,q,r\in R$ and quadratic equation has a no real roots that means $D<0$ and also as the value of p changes the sign then function also changes the sign and there are two conditions $p>0$ and $p<0$ solve this condition and you will get the answer.

Complete step-by-step solution:
It has mentioned in the question that $p,q,r\in R$ and the quadratic equation is given $p{{x}^{2}}+qx+r=0$ we also know that that if no real roots that means $D<0$
Where $D={{B}^{2}}-4AC$ substitute in this condition we get:
${{B}^{2}}-4AC<0$
$B=q$,$A=p$,$C=r$ Substitute this value on above equation we get:
${{q}^{2}}-4pr<0$
After simplifying this we get:
${{q}^{2}}<4pr$
Given function is $f(x)=p{{x}^{2}}+qx+r=0$
For no real roots value of p has two values positive as well as negative
Since the quadratic equation $f(x)=0$ has no real root, $f(x)$ will have the same sign as that of p for all $x\in R$
That means $p>0$ and $p<0$
For $p>0$
Now, we have to substitute the value of $x=1$ in this function
$f(1)=p+q+r$ But $f(1)>0$ therefore, $f(1)=p+q+r>0$
That means $f(p+q+r)>0$ which has no real roots
$pf(1)>0$ Because value of p and $f(1)$ are both positive
Other conditions is that
For$p<0$
Now, we have to substitute the value of $x=1$ in this function
$f(1)=p+q+r$ But $f(1)<0$ therefore, $f(1)=p+q+r<0$
That means $f(p+q+r)<0$ which has no real roots.
$pf(1)>0$ Because value of p and $f(1)$ are both negative in this case
Also, $f(0).f(1)>0$
$\therefore r(p+q+r)>0$
So, the correct option is “option 2”.

Note: In this particular problem we have to keep in mind that the sign of p changes with its function. Solve this problem step wise step to avoid the silly mistake and don’t confuse yourself among the conditions. So, the above solution is preferred to such types of problems.