Answer
414.6k+ views
Hint: First of all take all the terms of the given quadratic equation to the left so that we can get 0 in the right hand side. Now, assume the two roots of the given quadratic equation as \[\alpha \] and \[\beta \] with \[\alpha =2\]. Apply the identity: - \[\alpha +\beta =\dfrac{-b}{a}\] and \[\alpha \beta =\dfrac{c}{a}\] to form two linear equation in k and \[\beta \] and find the value of k. Remember that b = co-efficient of x and c = constant term while a = co – efficient of \[{{x}^{2}}\].
Complete step-by-step solution
We have been given the quadratic equation: -
\[\Rightarrow \]\[3{{x}^{2}}-9x=kx-k\]
Taking all the terms to the L.H.S, we get,
\[\begin{align}
& \Rightarrow 3{{x}^{2}}-9x-kx+k=0 \\
& \Rightarrow 3{{x}^{2}}-x\left( 9+k \right)+k=0 \\
\end{align}\]
Let us assume the two roots of the above equation as \[\alpha \] and \[\beta \]. We have provided with one root equal to 2. Let \[\alpha =2\]. Here, we are assuming the coefficient of \[{{x}^{2}}\], x, and the constant term is denoted by a, b and c respectively.
\[\Rightarrow a=3,b=-\left( 9+k \right)\] and $c = k.$
Now, we know that sum of the roots of a quadratic equation is the ratio of (-b) and (a) and product of the roots is the ratio of (c) and (a).
\[\begin{align}
& \Rightarrow \alpha +\beta =\dfrac{-b}{a} \\
& \Rightarrow 2+\beta =\dfrac{9+k}{3} \\
& \Rightarrow 6+3\beta =9+k \\
\end{align}\]
\[\Rightarrow 3\beta -k=3\] ----- (i)
Also, we have,
\[\begin{align}
& \Rightarrow \alpha \beta =\dfrac{c}{a} \\
& \Rightarrow 2\beta =\dfrac{k}{3} \\
\end{align}\]
\[\Rightarrow \beta =\dfrac{k}{6}\] ------ (ii)
Substituting the value of \[\beta \] from equation (ii) in equation (i), we have,
\[\begin{align}
& \Rightarrow 3\times \dfrac{k}{6}-k=3 \\
& \Rightarrow \dfrac{k}{2}-k=3 \\
& \Rightarrow -\dfrac{k}{2}=3 \\
& \Rightarrow k=-6 \\
\end{align}\]
Hence, option (c) is the correct answer.
Note: You may note that we have assumed \[\alpha =2\]. You may assume, \[\beta =2\], this will not affect the value of k in the answer. The only difference it will make is that at last, we have to eliminate \[\alpha \] instead of \[\beta \] just like we did above. There can be another way to solve the question if options are given. We will substitute the value of k one – by – one and find the roots. The option which will give one of the roots as 2 will be the answer.
Complete step-by-step solution
We have been given the quadratic equation: -
\[\Rightarrow \]\[3{{x}^{2}}-9x=kx-k\]
Taking all the terms to the L.H.S, we get,
\[\begin{align}
& \Rightarrow 3{{x}^{2}}-9x-kx+k=0 \\
& \Rightarrow 3{{x}^{2}}-x\left( 9+k \right)+k=0 \\
\end{align}\]
Let us assume the two roots of the above equation as \[\alpha \] and \[\beta \]. We have provided with one root equal to 2. Let \[\alpha =2\]. Here, we are assuming the coefficient of \[{{x}^{2}}\], x, and the constant term is denoted by a, b and c respectively.
\[\Rightarrow a=3,b=-\left( 9+k \right)\] and $c = k.$
Now, we know that sum of the roots of a quadratic equation is the ratio of (-b) and (a) and product of the roots is the ratio of (c) and (a).
\[\begin{align}
& \Rightarrow \alpha +\beta =\dfrac{-b}{a} \\
& \Rightarrow 2+\beta =\dfrac{9+k}{3} \\
& \Rightarrow 6+3\beta =9+k \\
\end{align}\]
\[\Rightarrow 3\beta -k=3\] ----- (i)
Also, we have,
\[\begin{align}
& \Rightarrow \alpha \beta =\dfrac{c}{a} \\
& \Rightarrow 2\beta =\dfrac{k}{3} \\
\end{align}\]
\[\Rightarrow \beta =\dfrac{k}{6}\] ------ (ii)
Substituting the value of \[\beta \] from equation (ii) in equation (i), we have,
\[\begin{align}
& \Rightarrow 3\times \dfrac{k}{6}-k=3 \\
& \Rightarrow \dfrac{k}{2}-k=3 \\
& \Rightarrow -\dfrac{k}{2}=3 \\
& \Rightarrow k=-6 \\
\end{align}\]
Hence, option (c) is the correct answer.
Note: You may note that we have assumed \[\alpha =2\]. You may assume, \[\beta =2\], this will not affect the value of k in the answer. The only difference it will make is that at last, we have to eliminate \[\alpha \] instead of \[\beta \] just like we did above. There can be another way to solve the question if options are given. We will substitute the value of k one – by – one and find the roots. The option which will give one of the roots as 2 will be the answer.
Recently Updated Pages
How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Why Are Noble Gases NonReactive class 11 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let X and Y be the sets of all positive divisors of class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x and y be 2 real numbers which satisfy the equations class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x 4log 2sqrt 9k 1 + 7 and y dfrac132log 2sqrt5 class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x22ax+b20 and x22bx+a20 be two equations Then the class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Trending doubts
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
At which age domestication of animals started A Neolithic class 11 social science CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Which are the Top 10 Largest Countries of the World?
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Give 10 examples for herbs , shrubs , climbers , creepers
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Difference Between Plant Cell and Animal Cell
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Write a letter to the principal requesting him to grant class 10 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Change the following sentences into negative and interrogative class 10 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Fill in the blanks A 1 lakh ten thousand B 1 million class 9 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)