# If one of the root of \[{\text{ }}2{x^2} - cx + 3 = 0{\text{ }}\]is 3 and the another equation \[2{x^2} - cx + d = 0\]has equal roots where c and d are real numbers, then d is equal to

\[

{\text{a}}{\text{. 3}} \\

{\text{b}}{\text{. }}\dfrac{{{\text{ 49}}}}{8} \\

{\text{c}}{\text{. }}\dfrac{8}{{49}} \\

{\text{d}}{\text{. }} - 3 \\

\]

Last updated date: 26th Mar 2023

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Answer

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Hint- In quadratic equation the sum of the roots is${\text{ = }}\left( {\dfrac{{ - {\text{coefficient of }}x}}{{{\text{coefficient of }}{x^2}}}} \right)$, and the product of the roots ${\text{ = }}\left( {\dfrac{{{\text{constant term}}}}{{{\text{coefficient of }}{x^2}}}} \right)$

As we know in quadratic equation the sum of the roots is ${\text{ = }}\left( {\dfrac{{ - {\text{coefficient of }}x}}{{{\text{coefficient of }}{x^2}}}} \right)$

So, first equation is \[{\text{ }}2{x^2} - cx + 3 = 0{\text{ }}\]

Let the roots of this equation be $\alpha ,{\text{ }}\beta $

It is given that one of the root is 3

So, let $\alpha {\text{ = 3}}$

$ \Rightarrow $Sum of the roots${\text{ = }}\alpha {\text{ + }}\beta {\text{ = }}\left( {\dfrac{{ - {\text{coefficient of }}x}}{{{\text{coefficient of }}{x^2}}}} \right) = \dfrac{c}{2}$

$ \Rightarrow 3 + \beta = \dfrac{c}{2} \Rightarrow \beta = \dfrac{c}{2} - 3................\left( 1 \right)$

And the product of the roots $\alpha \beta {\text{ = }}\left( {\dfrac{{{\text{constant term}}}}{{{\text{coefficient of }}{x^2}}}} \right) = \dfrac{3}{2}$

$ \Rightarrow 3\beta = \dfrac{3}{2} \Rightarrow \beta = \dfrac{1}{2}$

So, from equation 1

$

\beta = \dfrac{c}{2} - 3 \Rightarrow \dfrac{1}{2} = \dfrac{c}{2} - 3 \Rightarrow \dfrac{c}{2} = \dfrac{7}{2} \\

\Rightarrow c = 7................\left( 2 \right) \\

$

Another given equation is \[2{x^2} - cx + d = 0\]

It is given its roots are equal

So let its roots are $\lambda ,\lambda $

$ \Rightarrow $Sum of the roots${\text{ = }}\lambda {\text{ + }}\lambda {\text{ = }}\left( {\dfrac{{ - {\text{coefficient of }}x}}{{{\text{coefficient of }}{x^2}}}} \right) = \dfrac{c}{2}$

$ \Rightarrow 2\lambda = \dfrac{c}{2}$

From equation (2)

$ \Rightarrow 2\lambda = \dfrac{7}{2} \Rightarrow \lambda = \dfrac{7}{4}$

And the product of the roots $\lambda \lambda {\text{ = }}\left( {\dfrac{{{\text{constant term}}}}{{{\text{coefficient of }}{x^2}}}} \right) = \dfrac{d}{2}$

$

\Rightarrow {\lambda ^2} = \dfrac{d}{2} \Rightarrow {\left( {\dfrac{7}{4}} \right)^2} = \dfrac{d}{2} \\

\Rightarrow \dfrac{d}{2} = \dfrac{{49}}{{16}} \Rightarrow d = \dfrac{{49}}{8} \\

$

So, option (b) is correct.

Note- In such types of questions the key concept we have to remember is that always remember the sum and product of roots of the quadratic equation which is stated above, then apply these formulas in the given equations and after simplification we will get the required answer.

As we know in quadratic equation the sum of the roots is ${\text{ = }}\left( {\dfrac{{ - {\text{coefficient of }}x}}{{{\text{coefficient of }}{x^2}}}} \right)$

So, first equation is \[{\text{ }}2{x^2} - cx + 3 = 0{\text{ }}\]

Let the roots of this equation be $\alpha ,{\text{ }}\beta $

It is given that one of the root is 3

So, let $\alpha {\text{ = 3}}$

$ \Rightarrow $Sum of the roots${\text{ = }}\alpha {\text{ + }}\beta {\text{ = }}\left( {\dfrac{{ - {\text{coefficient of }}x}}{{{\text{coefficient of }}{x^2}}}} \right) = \dfrac{c}{2}$

$ \Rightarrow 3 + \beta = \dfrac{c}{2} \Rightarrow \beta = \dfrac{c}{2} - 3................\left( 1 \right)$

And the product of the roots $\alpha \beta {\text{ = }}\left( {\dfrac{{{\text{constant term}}}}{{{\text{coefficient of }}{x^2}}}} \right) = \dfrac{3}{2}$

$ \Rightarrow 3\beta = \dfrac{3}{2} \Rightarrow \beta = \dfrac{1}{2}$

So, from equation 1

$

\beta = \dfrac{c}{2} - 3 \Rightarrow \dfrac{1}{2} = \dfrac{c}{2} - 3 \Rightarrow \dfrac{c}{2} = \dfrac{7}{2} \\

\Rightarrow c = 7................\left( 2 \right) \\

$

Another given equation is \[2{x^2} - cx + d = 0\]

It is given its roots are equal

So let its roots are $\lambda ,\lambda $

$ \Rightarrow $Sum of the roots${\text{ = }}\lambda {\text{ + }}\lambda {\text{ = }}\left( {\dfrac{{ - {\text{coefficient of }}x}}{{{\text{coefficient of }}{x^2}}}} \right) = \dfrac{c}{2}$

$ \Rightarrow 2\lambda = \dfrac{c}{2}$

From equation (2)

$ \Rightarrow 2\lambda = \dfrac{7}{2} \Rightarrow \lambda = \dfrac{7}{4}$

And the product of the roots $\lambda \lambda {\text{ = }}\left( {\dfrac{{{\text{constant term}}}}{{{\text{coefficient of }}{x^2}}}} \right) = \dfrac{d}{2}$

$

\Rightarrow {\lambda ^2} = \dfrac{d}{2} \Rightarrow {\left( {\dfrac{7}{4}} \right)^2} = \dfrac{d}{2} \\

\Rightarrow \dfrac{d}{2} = \dfrac{{49}}{{16}} \Rightarrow d = \dfrac{{49}}{8} \\

$

So, option (b) is correct.

Note- In such types of questions the key concept we have to remember is that always remember the sum and product of roots of the quadratic equation which is stated above, then apply these formulas in the given equations and after simplification we will get the required answer.

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