Question

Given \[{{x}_{1}},{{x}_{2}}\] are the roots of the equation \[{{x}^{2}}-3x+A=0,\] and \[{{x}_{3}},{{x}_{4}}\] are the roots of the equation \[{{x}^{2}}-12x+B=0,\] such that \[{{x}_{1}},{{x}_{2}},{{x}_{3}},{{x}_{4}}\] form an increasing GP. Then find the value of A + B.

Answer 1

Hint: Assume \[{{x}_{1}},{{x}_{2}},{{x}_{3}},{{x}_{4}}\] as the general terms of the increasing GP as \[a,ar,a{{r}^{2}},a{{r}^{3}}\] respectively. Now, form two relations between ‘a’ and ‘r’ using the formula, \[\text{sum of roots}=\dfrac{-\text{ coefficient of x}}{\text{coefficient of }{{x}^{2}}}\] and find the value of ‘a’ and ‘r’. Here ‘a’ is the first term of the GP and ‘r’ is the common ratio. Now, use the formula, \[\text{product of roots}=\dfrac{\text{constant term}}{\text{coefficient of }{{x}^{2}}}\] and form the expressions of A and B in terms of \[{{x}_{1}},{{x}_{2}},{{x}_{3}},{{x}_{4}}\] and add them to get the value of A + B.

Complete step by step answer:
Here, we have been given two quadratic equations \[{{x}^{2}}-3x+A=0\] and \[{{x}^{2}}-12x+B=0\] with \[{{x}_{1}},{{x}_{2}}\] as the roots of the first equation and \[{{x}_{3}},{{x}_{4}}\] as the roots of the second equation. It is said that \[{{x}_{1}},{{x}_{2}},{{x}_{3}},{{x}_{4}}\] are forming an increasing GP and we have to find the value of A + B.
Now, let us assume \[{{x}_{1}},{{x}_{2}},{{x}_{3}},{{x}_{4}}\] as four general terms of GP as \[a,a{{r}^{1}},a{{r}^{2}},a{{r}^{3}}\] respectively. Here, ‘a’ is the first term and ‘r’ is the common ratio of the GP.
Now, we know that, \[\text{sum of roots}=\dfrac{-\text{ coefficient of x}}{\text{coefficient of }{{x}^{2}}}\] so, we have,
Considering the equation \[{{x}^{2}}-3x+A=0\] we have,
\[\Rightarrow {{x}_{1}}+{{x}_{2}}=\dfrac{-\left( -3 \right)}{1}\]
\[\Rightarrow a+ar=3\]
\[\Rightarrow a\left( 1+r \right)=3.....\left( i \right)\]
Considering the equation \[{{x}^{2}}-12x+B=0\] we have,
\[\Rightarrow {{x}_{3}}+{{x}_{4}}=12\]
\[\Rightarrow a{{r}^{2}}+a{{r}^{3}}=12\]
\[\Rightarrow a{{r}^{2}}\left( 1+r \right)=12.....\left( ii \right)\]
Dividing equations (ii) by (i), we get,
\[\Rightarrow \dfrac{a{{r}^{2}}\left( 1+r \right)}{a\left( 1+r \right)}=\dfrac{12}{3}\]
Cancelling the common terms, we get,
\[\Rightarrow {{r}^{2}}=4\]
\[\Rightarrow r=\pm 2\]
Here, we are neglecting r = – 2 because it is given that the GP has increasing terms. So, we have r = 2. Now, we know that the value of r is 2, so substituting r = 2 in equation (i), we get,
\[\Rightarrow a\times \left( 1+2 \right)=3\]
\[\Rightarrow a\times 3=3\]
\[\Rightarrow a=1\]
Therefore, the terms of GP are
\[{{x}_{1}}=a=1\]
\[{{x}_{2}}=ar=2\]
\[{{x}_{3}}=a{{r}^{2}}=4\]
\[{{x}_{4}}=a{{r}^{3}}=8\]
Now, we know that, \[\text{product of roots}=\dfrac{\text{constant term}}{\text{coefficient of }{{x}^{2}}},\] so we have,
Considering the equation \[{{x}^{2}}-3x+A=0,\] we have,
\[\Rightarrow {{x}_{1}}{{x}_{2}}=\dfrac{A}{1}\]
\[\Rightarrow {{x}_{1}}{{x}_{2}}=A.....\left( iii \right)\]
Considering the equation \[{{x}^{2}}-12x+B=0,\] we have,
\[\Rightarrow {{x}_{3}}{{x}_{4}}=\dfrac{B}{1}\]
\[\Rightarrow {{x}_{3}}{{x}_{4}}=B.....\left( iv \right)\]
Adding equation (iii) and (iv), we get,
\[\Rightarrow {{x}_{1}}{{x}_{2}}+{{x}_{3}}{{x}_{4}}=A+B\]
\[\Rightarrow A+B={{x}_{1}}{{x}_{2}}+{{x}_{3}}{{x}_{4}}\]
Substituting the values of \[{{x}_{1}},{{x}_{2}},{{x}_{3}},{{x}_{4}}\] we get,
\[\Rightarrow A+B=\left( 1\times 2 \right)+\left( 4\times 8 \right)\]
\[\Rightarrow A+B=2+32\]
\[\Rightarrow A+B=34\]
Hence, the value of A + B is 34.

Note:
One must not assume \[{{x}_{1}},{{x}_{2}},{{x}_{3}},{{x}_{4}}\] as four variables because then we will need four relations to find their values and calculation will be more. So, always write them as general terms of GP. You must remember the formulas of the sum and product of the roots of the quadratic equation to solve the question. You may note that we have neglected r = – 2 in the above solution because if we will assume r = – 2 then we cannot form a GP with the given terms \[{{x}_{1}},{{x}_{2}},{{x}_{3}},{{x}_{4}}.\]