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
\[
{\text{a}}{\text{. 3}} \\
{\text{b}}{\text{. }}\dfrac{{{\text{ 49}}}}{8} \\
{\text{c}}{\text{. }}\dfrac{8}{{49}} \\
{\text{d}}{\text{. }} - 3 \\
\]
Answer
367.2k+ views
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.
Last updated date: 01st Oct 2023
•
Total views: 367.2k
•
Views today: 8.67k
Recently Updated Pages
What do you mean by public facilities

Paragraph on Friendship

Slogan on Noise Pollution

Disadvantages of Advertising

Prepare a Pocket Guide on First Aid for your School

10 Slogans on Save the Tiger

Trending doubts
How do you solve x2 11x + 28 0 using the quadratic class 10 maths CBSE

Fill the blanks with the suitable prepositions 1 The class 9 english CBSE

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE

Difference Between Plant Cell and Animal Cell

The equation xxx + 2 is satisfied when x is equal to class 10 maths CBSE

Who had given the title of Mahatma to Gandhi Ji A Bal class 10 social science CBSE

Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE

How many millions make a billion class 6 maths CBSE

Find the value of the expression given below sin 30circ class 11 maths CBSE
