
If $\alpha $ and $\beta$ are roots of the equation $ax^{2}+2bx+c=0$, then
$\sqrt{\dfrac{\alpha }{\beta }}+\sqrt{\dfrac{\beta }{\alpha }}$ is equal to
A. $\dfrac{2b}{c}$
B.$\dfrac{2b}{\sqrt{ac}}$
C. $\dfrac{-2b}{\sqrt{ac}}$
D. $\dfrac{-b}{\sqrt{ac}}$
Answer
162.6k+ views
Hint:The values of the variable that satisfy a quadratic equation are known as its roots. The "solutions" or "zeros" of the quadratic equation are other names for them. We need to first simplify the given equation whose value we need to find and then substitute the value of the sum of the roots and the product of the roots in the simplified equation.
Formula Used:
Sum of roots,
$\alpha +\beta =\dfrac{-b}{a}$
Product of the roots,
$\alpha \beta =\dfrac{c}{a}$
Complete step by step Solution:
The given equation is
$ax^{2}+2bx+c=0$ which is quadratic in nature.
Let the roots of the quadratic equation be $\alpha$and $\beta$
Now, substituting the values of respective coefficients in the formula of the sum of roots and product of roots, we get
$\alpha +\beta =\dfrac{-b}{a}=\dfrac{-2b}{a}$
Also,
$\alpha \beta =\dfrac{c}{a}$
We need to find
$\sqrt{\dfrac{\alpha }{\beta }}+\sqrt{\dfrac{\beta }{\alpha }}$
By taking lcm of the above equation
$\dfrac{\left ( \sqrt{\alpha ^{2}} \right )+\left ( \sqrt{\beta^{2}} \right )}{\sqrt{\alpha \beta }}$
$= \dfrac{\alpha +\beta }{\sqrt{\alpha \beta }}$
$=\dfrac{\dfrac{-2b}{a}}{\sqrt{\dfrac{c}{a}}}$
Therefore,
$\sqrt{\dfrac{\alpha }{\beta }}+\sqrt{\dfrac{\beta }{\alpha }}=\dfrac{-2b}{\sqrt{ac}}$
Hence, the correct option is (C).
Additional Information:We know that the sum of the roots of a quadratic equation is the negative coefficient of the second term of the generalized form which is linear in nature divided by the coefficient of the first term of the generalized form of the equation which is quadratic in nature.
Also, the product of the roots is the third term of the generalized form of the equation which is constant in nature divided by the coefficient of the first term of the generalized form of the
equation which is quadratic in nature.
Note: Students must make sure that the simplified version of the given equation is either in the form sum of roots or the product of roots or a combination of both of them. This needs to be done so that we have values to substitute which will give our ultimate result.
Formula Used:
Sum of roots,
$\alpha +\beta =\dfrac{-b}{a}$
Product of the roots,
$\alpha \beta =\dfrac{c}{a}$
Complete step by step Solution:
The given equation is
$ax^{2}+2bx+c=0$ which is quadratic in nature.
Let the roots of the quadratic equation be $\alpha$and $\beta$
Now, substituting the values of respective coefficients in the formula of the sum of roots and product of roots, we get
$\alpha +\beta =\dfrac{-b}{a}=\dfrac{-2b}{a}$
Also,
$\alpha \beta =\dfrac{c}{a}$
We need to find
$\sqrt{\dfrac{\alpha }{\beta }}+\sqrt{\dfrac{\beta }{\alpha }}$
By taking lcm of the above equation
$\dfrac{\left ( \sqrt{\alpha ^{2}} \right )+\left ( \sqrt{\beta^{2}} \right )}{\sqrt{\alpha \beta }}$
$= \dfrac{\alpha +\beta }{\sqrt{\alpha \beta }}$
$=\dfrac{\dfrac{-2b}{a}}{\sqrt{\dfrac{c}{a}}}$
Therefore,
$\sqrt{\dfrac{\alpha }{\beta }}+\sqrt{\dfrac{\beta }{\alpha }}=\dfrac{-2b}{\sqrt{ac}}$
Hence, the correct option is (C).
Additional Information:We know that the sum of the roots of a quadratic equation is the negative coefficient of the second term of the generalized form which is linear in nature divided by the coefficient of the first term of the generalized form of the equation which is quadratic in nature.
Also, the product of the roots is the third term of the generalized form of the equation which is constant in nature divided by the coefficient of the first term of the generalized form of the
equation which is quadratic in nature.
Note: Students must make sure that the simplified version of the given equation is either in the form sum of roots or the product of roots or a combination of both of them. This needs to be done so that we have values to substitute which will give our ultimate result.
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