
If A.M of the roots of a quadratic equation is $\dfrac{8}{5}$ and A.M. of their reciprocals is $\dfrac{8}{7}$, then the equation is?
( a ) $7{{x}^{2}}-16x+8=0$
( b ) $3{{x}^{2}}-12x+7=0$
( c ) $5{{x}^{2}}-16x+7=0$
( d ) $7{{x}^{2}}-16x+5=0$
Answer
161.1k+ views
Hint: In this question, we are the arithmetic mean of the quadratic equation and their reciprocals and we have to find the quadratic equation. To solve this, we suppose the roots be $\alpha $ and $\beta $, and by using the formula of A.M and putting the value we find the sum of roots. Similarly, by reciprocal it and putting the values, we find the product of roots, and then putting the values in the standard quadratic equation, we are able to get the quadratic equation.
Formula Used: Arithmetic mean = $\dfrac{Sum\, of\,values}{Number\,of\,values}$
Complete step by step Solution:
Given that A.M. of the roots of a quadratic equation is $\dfrac{8}{5}$and their reciprocals is $\dfrac{8}{7}$
We have to find the quadratic equation.
Suppose that $\alpha $ and $\beta $ are the roots of the quadratic equation
According to the question
$\dfrac{\alpha +\beta }{2}=\dfrac{8}{5}$
Then $\alpha +\beta =\dfrac{16}{5}$
And their reciprocals is $\dfrac{8}{7}$
Then $\dfrac{\dfrac{1}{\alpha }+\dfrac{1}{\beta }}{2}=\dfrac{8}{7}$
That is $\dfrac{\alpha +\beta }{2\alpha \beta }=\dfrac{8}{7}$
By solving further, we get
$7(\alpha +\beta )=16\alpha \beta $
By pitting the value of $\alpha +\beta $ in the above equation, we get
$\dfrac{7\times \dfrac{16}{5}}{16}=\alpha \beta $
That is $\alpha \beta =\dfrac{7}{5}$
Now we know quadratic equation with its roots is
${{x}^{2}}-(\alpha +\beta )x+(\alpha \beta )=0$
By putting the values of $\alpha +\beta $and $\alpha \beta $in the above equation , we get
${{x}^{2}}-\dfrac{16}{5}x+\dfrac{7}{5}=0$
That is $5{{x}^{2}}-16x+7=0$
Therefore, the correct option is (c).
Note: Arithmetic mean is simply adding the values and dividing it with number of roots. As we know in quadratic equation, we get two roots by solving it. So we suppose the two roots be $\alpha $ and $\beta $ and divide it by 2 and the arithmetic mean is given in the question. By putting the values, we are able to solve it.
Formula Used: Arithmetic mean = $\dfrac{Sum\, of\,values}{Number\,of\,values}$
Complete step by step Solution:
Given that A.M. of the roots of a quadratic equation is $\dfrac{8}{5}$and their reciprocals is $\dfrac{8}{7}$
We have to find the quadratic equation.
Suppose that $\alpha $ and $\beta $ are the roots of the quadratic equation
According to the question
$\dfrac{\alpha +\beta }{2}=\dfrac{8}{5}$
Then $\alpha +\beta =\dfrac{16}{5}$
And their reciprocals is $\dfrac{8}{7}$
Then $\dfrac{\dfrac{1}{\alpha }+\dfrac{1}{\beta }}{2}=\dfrac{8}{7}$
That is $\dfrac{\alpha +\beta }{2\alpha \beta }=\dfrac{8}{7}$
By solving further, we get
$7(\alpha +\beta )=16\alpha \beta $
By pitting the value of $\alpha +\beta $ in the above equation, we get
$\dfrac{7\times \dfrac{16}{5}}{16}=\alpha \beta $
That is $\alpha \beta =\dfrac{7}{5}$
Now we know quadratic equation with its roots is
${{x}^{2}}-(\alpha +\beta )x+(\alpha \beta )=0$
By putting the values of $\alpha +\beta $and $\alpha \beta $in the above equation , we get
${{x}^{2}}-\dfrac{16}{5}x+\dfrac{7}{5}=0$
That is $5{{x}^{2}}-16x+7=0$
Therefore, the correct option is (c).
Note: Arithmetic mean is simply adding the values and dividing it with number of roots. As we know in quadratic equation, we get two roots by solving it. So we suppose the two roots be $\alpha $ and $\beta $ and divide it by 2 and the arithmetic mean is given in the question. By putting the values, we are able to solve it.
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