Question

Find a quadratic equation whose product of roots \[{{x}_{1}}\]and \[{{x}_{2}}\] is equal to 4 and satisfying \[\dfrac{{{x}_{1}}}{{{x}_{1}}-1}+\dfrac{{{x}_{2}}}{{{x}_{2}}-1}=2\].
\[\begin{align}
  & \text{(A) }{{\text{x}}^{2}}-2x+4=0 \\
 & (\text{B) }{{\text{x}}^{2}}-3x+4=0 \\
 & (\text{C) }{{\text{x}}^{2}}-2x+3=0 \\
 & \text{(D) }{{\text{x}}^{2}}-3x+2=0 \\
\end{align}\]

Answer 1

Hint: Let us assume the roots of \[a{{x}^{2}}+bx+c=0\] are \[{{x}_{1}}\] and \[{{x}_{2}}\].In the question we were given that the product of roots \[{{x}_{1}}\]and \[{{x}_{2}}\] is equal to 4. Let us assume this equation as equation (1). Now by using the relation \[\dfrac{{{x}_{1}}}{{{x}_{1}}-1}+\dfrac{{{x}_{2}}}{{{x}_{2}}-1}=2\], we should find the relation between \[{{x}_{1}}\] and \[{{x}_{2}}\]. Let us assume this equation as equation (2). By solving equation (1) and equation (2), we can get the quadratic equation.

Complete step-by-step answer:
Before solving the question, we should know that if \[\alpha \] and \[\beta \] are the roots of \[a{{x}^{2}}+bx+c=0\], then \[\alpha +\beta =\dfrac{-b}{a}\] and \[\alpha \beta =\dfrac{c}{a}\].
Let us assume a quadratic equation \[a{{x}^{2}}+bx+c=0\] whose roots are \[{{x}_{1}}\]and \[{{x}_{2}}\].
In the question, we are given that the product of roots \[{{x}_{1}}\]and \[{{x}_{2}}\] is equal to 4.
\[\Rightarrow {{x}_{1}}{{x}_{2}}=4.....(1)\]
In the question, we were also given \[\dfrac{{{x}_{1}}}{{{x}_{1}}-1}+\dfrac{{{x}_{2}}}{{{x}_{2}}-1}=2\]
\[\begin{align}
  & \Rightarrow \dfrac{{{x}_{1}}({{x}_{2}}-1)+{{x}_{2}}({{x}_{1}}-1)}{({{x}_{1}}-1)({{x}_{2}}-1)}=2 \\
 & \Rightarrow \dfrac{{{x}_{1}}{{x}_{2}}-{{x}_{1}}+{{x}_{1}}{{x}_{2}}-{{x}_{2}}}{{{x}_{1}}{{x}_{2}}-{{x}_{2}}-{{x}_{1}}+1}=2 \\
 & \Rightarrow \dfrac{2{{x}_{1}}{{x}_{2}}-({{x}_{1}}+{{x}_{2}})}{{{x}_{1}}{{x}_{2}}-({{x}_{1}}+{{x}_{2}})+1}=2 \\
\end{align}\]
By using cross multiplication, we get
\[\begin{align}
  & \Rightarrow 2{{x}_{1}}{{x}_{2}}-({{x}_{1}}+{{x}_{2}})=2\left( {{x}_{1}}{{x}_{2}}-({{x}_{1}}+{{x}_{2}})+1 \right) \\
 & \Rightarrow 2{{x}_{1}}{{x}_{2}}-({{x}_{1}}+{{x}_{2}})=2{{x}_{1}}{{x}_{2}}-2({{x}_{1}}+{{x}_{2}})+2 \\
 & \Rightarrow ({{x}_{1}}+{{x}_{2}})=2.....(4) \\
\end{align}\]
From equation (3), we get that the sum of roots of \[a{{x}^{2}}+bx+c=0\] is equal to 2.
From equation (4), we get that the product of roots of \[a{{x}^{2}}+bx+c=0\]is equal 4.
We know that if the sum of roots of a quadratic equation is \[{{x}_{1}}+{{x}_{2}}\] and product of roots of a quadratic equation is \[{{x}_{1}}{{x}_{2}}\], then the quadratic equation is \[{{x}^{2}}-({{x}_{1}}+{{x}_{2}})x+{{x}_{1}}{{x}_{2}}=0\].
So, the quadratic equation whose sum of roots are 2 and product of roots are 4 is equal to \[{{x}^{2}}-2x+4=0\].
So, the required quadratic equation is equal to \[{{x}^{2}}-2x+4=0\].
Hence, option (A) is correct.

Note: We can find the roots of \[a{{x}^{2}}+bx+c=0\] by using the formula \[x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\].
By using this formula, we can find the roots of \[{{x}^{2}}-2x+4=0\].
Now by comparing \[{{x}^{2}}-2x+4=0\] with \[a{{x}^{2}}+bx+c=0\], we get a=1, b=-2 and c=4.
So, the roots of equation of \[{{x}^{2}}-2x+4=0\] is equal to
\[x=\dfrac{2\pm \sqrt{4-4(4)}}{2}=\dfrac{2\pm \sqrt{-12}}{2}=\dfrac{2\pm i\sqrt{12}}{2}=\dfrac{2\pm 2i\sqrt{3}}{2}=1\pm i\sqrt{3}({{i}^{2}}=-1)\]
Therefore, we get that the roots of \[{{x}^{2}}-2x+4=0\] are \[1\pm i\sqrt{3}\]. So now we can check the conditions and confirm.