(C) Let the roots of the quadratic equation be $a$ and $b$. Given that the arithmetic mean of $a$ and $b$ is $34$,we have $\frac{a+b}{2} = 34$,which i

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(C) Let the roots of the quadratic equation be $a$ and $b$. Given that the arithmetic mean of $a$ and $b$ is $34$,we have $\frac{a+b}{2} = 34$,which implies $a+b = 68$. Given that the geometric mean of $a$ and $b$ is $16$,we have $\sqrt{ab} = 16$,which implies $ab = 16^{2} = 256$. $A$ quadratic equation with roots $a$ and $b$ is given by $x^{2} - (a+b)x + ab = 0$. Substituting the values,we get $x^{2} - 68x + 256 = 0$.

Class 11 Mathematics 4-2.Quadratic Equations and Inequations Solution of quadratic equations and Nature of roots

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The quadratic equation whose roots are the numbers having arithmetic mean $34$ and geometric mean $16$ is

MHT CET 2020 All MHT CET

A
$x^{2}+68x-256=0$
B
$x^{2}-68x-256=0$
C
$x^{2}-68x+256=0$
D
$x^{2}+68x+256=0$

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4-2.Quadratic Equations and Inequations

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