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Question: Let \(\alpha \) and \(\beta \) are the roots of the equation \(p{{x}^{2}}+qx+r=0\) , \(p\ne 0\). If ...

Let α\alpha and β\beta are the roots of the equation px2+qx+r=0p{{x}^{2}}+qx+r=0 , p≠0p\ne 0. If pp , qq and rr are in AP and 1α+1β=4\dfrac{1}{\alpha }+\dfrac{1}{\beta }=4 , then the value of ∣α−β∣\left| \alpha -\beta \right| is

  1. 619\dfrac{\sqrt{61}}{9}
  2. 2179\dfrac{2\sqrt{17}}{9}
  3. 349\dfrac{\sqrt{34}}{9}
  4. 2139\dfrac{2\sqrt{13}}{9}
Explanation

Solution

In this problem we need to find the value of ∣α−β∣\left| \alpha -\beta \right|. For this we will first consider the given quadratic equation px2+qx+r=0p{{x}^{2}}+qx+r=0 and the roots α\alpha and β\beta , use the relation between the roots and coefficients of the quadratic equation and write the values of α+β\alpha +\beta and αβ\alpha \beta . Now consider the statement that pp , qq and rr are in AP. From this we can establish a relation between pp , qq and rr. Now consider the value 1α+1β=4\dfrac{1}{\alpha }+\dfrac{1}{\beta }=4 apply the lcm and use the values of α+β\alpha +\beta and αβ\alpha \beta to get the relation between pp , qq and rr based on these relations we can calculate the required value.

Complete step-by-step solution:
The quadratic equation is px2+qx+r=0p{{x}^{2}}+qx+r=0 and the roots of the equation are α\alpha and β\beta .
We can write the values of α+β\alpha +\beta and αβ\alpha \beta as
α+β=−qp\alpha +\beta =-\dfrac{q}{p} and αβ=rp\alpha \beta =\dfrac{r}{p} .
Given that pp , qq and rr are in AP, we can write that 2q=p+r2q=p+r .
Also we have the value 1α+1β=4\dfrac{1}{\alpha }+\dfrac{1}{\beta }=4. Simplify the above value using LCM, then we will have
α+βαβ=4 α+β=4αβ \begin{aligned} & \dfrac{\alpha +\beta }{\alpha \beta }=4 \\\ & \alpha +\beta =4\alpha \beta \\\ \end{aligned}
Substitute the values α+β=−qp\alpha +\beta =-\dfrac{q}{p} and αβ=rp\alpha \beta =\dfrac{r}{p} in the above equation, then we will get
−qp=4rp ⇒q=−4r \begin{aligned} & -\dfrac{q}{p}=\dfrac{4r}{p} \\\ & \Rightarrow q=-4r \\\ \end{aligned}
We have the relation 2q=p+r2q=p+r. Substituting the value q=−4rq=-4r in this relation, then we will have
2(−4r)=p+r ⇒−8r=p+r ⇒p=−9r \begin{aligned} & 2\left( -4r \right)=p+r \\\ & \Rightarrow -8r=p+r \\\ & \Rightarrow p=-9r \\\ \end{aligned}
Now consider the value α+β=−qp\alpha +\beta =-\dfrac{q}{p}. Substitute the value of qq and pp , then we will get
α+β=−(−4r)(−9r) ⇒α+β=−49 \begin{aligned} & \alpha +\beta =-\dfrac{\left( -4r \right)}{\left( -9r \right)} \\\ & \Rightarrow \alpha +\beta =-\dfrac{4}{9} \\\ \end{aligned}
Consider the value αβ=rp\alpha \beta =\dfrac{r}{p}. Substitute the value of pp, then we will have
αβ=r−9r ⇒αβ=−19 \begin{aligned} & \alpha \beta =\dfrac{r}{-9r} \\\ & \Rightarrow \alpha \beta =-\dfrac{1}{9} \\\ \end{aligned}
From algebraic formulas (a+b)2=a2+b2+2ab{{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab and (a−b)2=a2+b2−2ab{{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab, we can write the value (α−β)2{{\left( \alpha -\beta \right)}^{2}} as
(α−β)2=(α+β)2−4αβ{{\left( \alpha -\beta \right)}^{2}}={{\left( \alpha +\beta \right)}^{2}}-4\alpha \beta
Substituting the values of α+β\alpha +\beta and αβ\alpha \beta in the above equation, then we will have
(α−β)2=(−49)2−4(−19) ⇒(α−β)2=1681+49 ⇒(α−β)2=16+9(4)81 ⇒(α−β)2=5281 \begin{aligned} & {{\left( \alpha -\beta \right)}^{2}}={{\left( -\dfrac{4}{9} \right)}^{2}}-4\left( -\dfrac{1}{9} \right) \\\ & \Rightarrow {{\left( \alpha -\beta \right)}^{2}}=\dfrac{16}{81}+\dfrac{4}{9} \\\ & \Rightarrow {{\left( \alpha -\beta \right)}^{2}}=\dfrac{16+9\left( 4 \right)}{81} \\\ & \Rightarrow {{\left( \alpha -\beta \right)}^{2}}=\dfrac{52}{81} \\\ \end{aligned}
Applying square root on both sides of the above equation, then we will get
∣α−β∣=2139\left| \alpha -\beta \right|=\dfrac{2\sqrt{13}}{9}
Hence option 4 is the correct answer.

Note: For this problem we can also use different methods. In which we can directly use the algebraic formula (α−β)2=(α+β)2−4αβ{{\left( \alpha -\beta \right)}^{2}}={{\left( \alpha +\beta \right)}^{2}}-4\alpha \beta at the beginning and substitute the values of α+β\alpha +\beta and αβ\alpha \beta . After that use the remaining data and simplify the equation to get the direct result. But it will have some complicated calculations. So we have not used that method.