Question

If \[3{{p}^{2}}=5p+2\] and \[3{{q}^{2}}=5q+2\], where \[p\ne q\], then the equation, whose roots are \[3p-2q\] and \[3q-2p\], is
1. \[3{{x}^{2}}-5x-100=0\]
2. \[5{{x}^{2}}+3x+100=0\]
3. \[3{{x}^{2}}-5x+100=0\]
4. \[5{{x}^{2}}-3x-100=0\]

Answer 1

Hint: To solve such types of problems, first of all in the question there are two roots given \[\alpha =3p-2q\] and \[\beta =3q-2p\] . So, we have to find the value of \[p\] and \[q\] from the quadratic equation which is given in the question. Then we have to substitute the value in this equation \[{{x}^{2~}}-\left( \alpha \text{ }+\text{ }\beta \right)\text{ }x+\alpha \beta =0\].

Complete step by step answer:
There are two equation are given in the question:
\[3{{p}^{2}}=5p+2\]
After rearranging the terms we get:
\[3{{p}^{2}}-5p-2=0----(1)\]
Next equation is given by:
\[3{{q}^{2}}=5q+2\]
After rearranging the terms we get:
\[3{{q}^{2}}-5q-2=0----(2)\]
By solving the quadratic equation of equation \[(1)\] we get the value of \[p\]
\[=(3p+1)(p-2)=0\]
\[p=\dfrac{-1}{3}\,\,,\,\,p=2\]
By solving the quadratic equation of equation \[(2)\] we get the value of \[q\]
\[q=\dfrac{-1}{3}\,\,,\,\,q=2\]
But the condition which is given in the question \[p\ne q\]
That means \[p=2\] then \[q=\dfrac{-1}{3}\,\] or vice versa
But we have to find the equation whose roots are given below:
\[\alpha =3p-2q---(3)\]
\[\beta =3q-2p---(4)\]
To find the roots we have to substitute the values of \[p\] as well as \[q\].
Substitute the value of \[p=2\] and \[q=\dfrac{-1}{3}\,\] in equation \[(3)\]
\[\alpha =3\left( 2 \right)-2\left( \dfrac{-1}{3} \right)\]
After simplification we get:
\[\alpha =6+\dfrac{2}{3}\]
\[\alpha =\dfrac{20}{3}\]
Substitute the value of \[p=2\] and \[q=\dfrac{-1}{3}\,\] in equation \[(4)\]
\[\beta =3\left( \dfrac{-1}{3} \right)-2\left( 2 \right)\]
After simplification we get:
\[\beta =-1-4\]
\[\beta =-5\]
The standard required equation is that
\[~{{x}^{2~}}-\left( \alpha \text{ }+\text{ }\beta \right)\text{ }x+\alpha \beta =0---(5)\]
For that we have to find the value of \[\left( \alpha \text{ }+\text{ }\beta \right)\] and \[\alpha \beta \]
By substituting the values of \[\alpha =\dfrac{20}{3}\] and \[\beta =-5\] in \[\left( \alpha \text{ }+\text{ }\beta \right)\]
\[\therefore \alpha +\beta =\dfrac{20}{3}+(-5)\]
\[\alpha +\beta =\dfrac{20}{3}-5\]
Further simplification we get
\[\alpha +\beta =\dfrac{20-15}{3}\]
\[\alpha +\beta =\dfrac{5}{3}---(6)\]
\[\alpha .\beta =\left( \dfrac{20}{3} \right)(-5)\]
Further simplification we get:
\[\therefore \alpha .\beta =\dfrac{-100}{3}---(7)\]
After substituting the value of equation \[(6)\] and equation \[(7)\] in equation \[(5)\]
\[\therefore \,~{{x}^{2~}}-\left( \dfrac{5}{3} \right)\text{ }x+\left( \dfrac{-100}{3} \right)=0\]
Further simplification we get:
\[\therefore \,~{{x}^{2~}}-\left( \dfrac{5}{3} \right)\text{ }x-\left( \dfrac{100}{3} \right)=0\]
Multiply by \[~3\]throughout equation you get:
\[\therefore \,~3{{x}^{2~}}-5\text{ }x-100=0\]

So, the correct answer is “Option 1”.

Note: There are two equations which are mentioned in the question in that we have to find the value of p as well as q. while selecting the value of p and q to find the roots of \[\alpha \] and \[\beta \] we cannot select the value of \[p=\dfrac{-1}{3}\,\]and \[q=\dfrac{-1}{3}\] nor \[p=2\]and \[q=2\] because the condition which is given as \[p\ne q\] (don’t ignore this condition) this conditions means that value of p and q are different hence we should take the value as \[p=2\],\[q=\dfrac{-1}{3}\,\]. You can also select the value as \[p=\dfrac{-1}{3}\,\], \[p=2\] and you will get the same equation as in the above solution.