# If p, q, r and s are real numbers such that \[pr=2\left( q+s \right)\], then show that at least one of the equations \[{{x}^{2}}+px+q=0\] and \[{{x}^{2}}+rx+s=0\] has real roots.

Answer

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Hint: Roots of quadratic equation are real if and only if the discriminant i.e. $\Delta ={{b}^{2}}-4ac>0$, we will use this concept to solve the above problem.

Complete step-by-step answer:

Given quadratic equations are \[{{x}^{2}}+px+q=0\] and \[{{x}^{2}}+rx+s=0\] . Let us name them as equation (1) and equation (2) respectively.

Now we will find discriminants for equation (1) and equation (2).

The above quadratic equations have real roots, so we will find discriminants for equation (1) and equation (2) and then solve the question.

Now we will find the discriminant of equation (1) as b=p, a=1, c=q.

${{\Delta }_{1}}={{b}^{2}}-4ac={{p}^{2}}-4q$

The discriminant of equation (1) is ${{p}^{2}}-4q$ Let this be equation (3).

Now we will find the discriminant of equation (1) as b=r, a=1, c=s.

${{\Delta }_{2}}={{b}^{2}}-4ac={{r}^{2}}-4s$

The discriminant of equation (2) is ${{r}^{2}}-4s$ and let this be equation (4).

Now we will add equation (3) and equation (4) that is adding ${{\Delta }_{1}}+{{\Delta }_{2}}$ , we will get

$\Rightarrow {{\Delta }_{1}}+{{\Delta }_{2}}=({{p}^{2}}-4q)+({{r}^{2}}-4s)$.

Further simplifying it we will get,

$\Rightarrow {{\Delta }_{1}}+{{\Delta }_{2}}=({{p}^{2}}+{{r}^{2}}-4q-4s)$, Now we will take -2 common from $-4q-4s$ we will get,

$\Rightarrow {{\Delta }_{1}}+{{\Delta }_{2}}=(({{p}^{2}}+{{r}^{2}})-2(2q+2s))\cdot \cdot \cdot \cdot \cdot \left( 5 \right)$

We know that $2q+2s=pr$ , as it was given in the above question.

Now we will substitute it in equation 5.

We will get, $\Rightarrow {{\Delta }_{1}}+{{\Delta }_{2}}=(({{p}^{2}}+{{r}^{2}})-2(2q+2s))\cdot \cdot \cdot \cdot \cdot \left( 5 \right)$

$\Rightarrow {{\Delta }_{1}}+{{\Delta }_{2}}=(({{p}^{2}}+{{r}^{2}})-2(pr))$

We know a basic algebraic formula, ${{a}^{2}}+{{b}^{2}}-2ab={{(a-b)}^{2}}$ now we will use it in above equation.

$(({{p}^{2}}+{{r}^{2}})-2(pr))$ can be written as ${{(p-r)}^{2}}$.

So, $\Rightarrow {{\Delta }_{1}}+{{\Delta }_{2}}={{(p-r)}^{2}}$ which is always greater than or equals to 0.

So, ${{\Delta }_{1}}+{{\Delta }_{2}}\ge 0$

This means both ${{\Delta }_{1}}$ and ${{\Delta }_{2}}$ cannot be negative simultaneously which means at least one or both

${{\Delta }_{1}}$ and ${{\Delta }_{2}}$ should be greater than zero, which means at least one of the equations \[{{x}^{2}}+px+q=0\] and \[{{x}^{2}}+rx+s=0\] has real roots.

Note: Analysing is the most important thing that has to be done to solve a problem, basic algebraic formula has to be remembered and can be used wherever needed and the perfect square cannot be negative. There are three types of roots and their nature can be determined by $\Delta $. If $\Delta $ >0 roots are real and if $\Delta $ <0 roots are imaginary and $\Delta $ =0, roots are equal.

Complete step-by-step answer:

Given quadratic equations are \[{{x}^{2}}+px+q=0\] and \[{{x}^{2}}+rx+s=0\] . Let us name them as equation (1) and equation (2) respectively.

Now we will find discriminants for equation (1) and equation (2).

The above quadratic equations have real roots, so we will find discriminants for equation (1) and equation (2) and then solve the question.

Now we will find the discriminant of equation (1) as b=p, a=1, c=q.

${{\Delta }_{1}}={{b}^{2}}-4ac={{p}^{2}}-4q$

The discriminant of equation (1) is ${{p}^{2}}-4q$ Let this be equation (3).

Now we will find the discriminant of equation (1) as b=r, a=1, c=s.

${{\Delta }_{2}}={{b}^{2}}-4ac={{r}^{2}}-4s$

The discriminant of equation (2) is ${{r}^{2}}-4s$ and let this be equation (4).

Now we will add equation (3) and equation (4) that is adding ${{\Delta }_{1}}+{{\Delta }_{2}}$ , we will get

$\Rightarrow {{\Delta }_{1}}+{{\Delta }_{2}}=({{p}^{2}}-4q)+({{r}^{2}}-4s)$.

Further simplifying it we will get,

$\Rightarrow {{\Delta }_{1}}+{{\Delta }_{2}}=({{p}^{2}}+{{r}^{2}}-4q-4s)$, Now we will take -2 common from $-4q-4s$ we will get,

$\Rightarrow {{\Delta }_{1}}+{{\Delta }_{2}}=(({{p}^{2}}+{{r}^{2}})-2(2q+2s))\cdot \cdot \cdot \cdot \cdot \left( 5 \right)$

We know that $2q+2s=pr$ , as it was given in the above question.

Now we will substitute it in equation 5.

We will get, $\Rightarrow {{\Delta }_{1}}+{{\Delta }_{2}}=(({{p}^{2}}+{{r}^{2}})-2(2q+2s))\cdot \cdot \cdot \cdot \cdot \left( 5 \right)$

$\Rightarrow {{\Delta }_{1}}+{{\Delta }_{2}}=(({{p}^{2}}+{{r}^{2}})-2(pr))$

We know a basic algebraic formula, ${{a}^{2}}+{{b}^{2}}-2ab={{(a-b)}^{2}}$ now we will use it in above equation.

$(({{p}^{2}}+{{r}^{2}})-2(pr))$ can be written as ${{(p-r)}^{2}}$.

So, $\Rightarrow {{\Delta }_{1}}+{{\Delta }_{2}}={{(p-r)}^{2}}$ which is always greater than or equals to 0.

So, ${{\Delta }_{1}}+{{\Delta }_{2}}\ge 0$

This means both ${{\Delta }_{1}}$ and ${{\Delta }_{2}}$ cannot be negative simultaneously which means at least one or both

${{\Delta }_{1}}$ and ${{\Delta }_{2}}$ should be greater than zero, which means at least one of the equations \[{{x}^{2}}+px+q=0\] and \[{{x}^{2}}+rx+s=0\] has real roots.

Note: Analysing is the most important thing that has to be done to solve a problem, basic algebraic formula has to be remembered and can be used wherever needed and the perfect square cannot be negative. There are three types of roots and their nature can be determined by $\Delta $. If $\Delta $ >0 roots are real and if $\Delta $ <0 roots are imaginary and $\Delta $ =0, roots are equal.

Last updated date: 26th Sep 2023

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