
If $\alpha {\text{ and }}\beta {\text{ }}$ are the roots of the equation ${{\text{x}}^2} + 3{\text{x - 4 = 0,then }}\dfrac{1}{\alpha } + \dfrac{1}{\beta }{\text{ is equal to}}$
A. $\dfrac{{ - 3}}{4}$
B. $\dfrac{3}{4}$
C. $\dfrac{{ - 4}}{3}$
D. $\dfrac{4}{3}$
E. $\dfrac{3}{2}$
Answer
611.7k+ views
Hint: In this question, a quadratic equation is given whose roots are $\alpha {\text{ and }}\beta {\text{ }}$and we have to find the value of a expression given in terms of $\alpha {\text{ and }}\beta {\text{ }}$. First write the given expression in terms of $\alpha + \beta {\text{ and }}\alpha \beta $ and then use the relationship between the sum and product of roots and the coefficient of the quadratic equation.
Complete step-by-step answer:
In the question, it is given that $\alpha {\text{ and }}\beta {\text{ }}$are roots of the quadratic equation ${{\text{x}}^2} + 3{\text{x - 4 = 0}}$ .
We have to find the value of the expression $\dfrac{1}{\alpha } + \dfrac{1}{\beta }$ .
We know that the relationship between the sum and product of roots and the coefficients of the quadratic equation ${\text{a}}{{\text{x}}^2} + {\text{bx + c = 0}}$ is given as:
$
\alpha + \beta = \dfrac{{ - {\text{b}}}}{{\text{a}}}. \\
\alpha \beta = \dfrac{{\text{c}}}{a} \\
$
On comparing the given quadratic equation with the general expression of the quadratic equation, we get:
a = 1 , b = 3 and c = -4.
Now, we will write the expression $\dfrac{1}{\alpha } + \dfrac{1}{\beta }$ in terms of $\alpha + \beta {\text{ and }}\alpha \beta $ .
$\dfrac{1}{\alpha } + \dfrac{1}{\beta } = \dfrac{{\alpha + \beta }}{{\alpha \beta }}$
Also
$
\alpha + \beta = \dfrac{{ - {\text{b}}}}{{\text{a}}} = \dfrac{{ - 3}}{1} = - 3 \\
\alpha \beta = \dfrac{{\text{c}}}{a} = \dfrac{{ - 4}}{1} = - 4 \\
$
Putting the values of $\alpha + \beta {\text{ and }}\alpha \beta $ in above expression, we get:
$\dfrac{1}{\alpha } + \dfrac{1}{\beta } = \dfrac{{\alpha + \beta }}{{\alpha \beta }} = \dfrac{{ - 3}}{{ - 4}} = \dfrac{3}{4}$ .
So option B is correct.
Note: In this type of question, we should know the general expression of a quadratic equation in terms of a, b and c. We should also remember the relation between the sum and product of roots and the coefficients of the quadratic equation ${\text{a}}{{\text{x}}^2} + {\text{bx + c = 0}}$. Finally convert the given expression in the terms of $\alpha + \beta {\text{ and }}\alpha \beta $.
Complete step-by-step answer:
In the question, it is given that $\alpha {\text{ and }}\beta {\text{ }}$are roots of the quadratic equation ${{\text{x}}^2} + 3{\text{x - 4 = 0}}$ .
We have to find the value of the expression $\dfrac{1}{\alpha } + \dfrac{1}{\beta }$ .
We know that the relationship between the sum and product of roots and the coefficients of the quadratic equation ${\text{a}}{{\text{x}}^2} + {\text{bx + c = 0}}$ is given as:
$
\alpha + \beta = \dfrac{{ - {\text{b}}}}{{\text{a}}}. \\
\alpha \beta = \dfrac{{\text{c}}}{a} \\
$
On comparing the given quadratic equation with the general expression of the quadratic equation, we get:
a = 1 , b = 3 and c = -4.
Now, we will write the expression $\dfrac{1}{\alpha } + \dfrac{1}{\beta }$ in terms of $\alpha + \beta {\text{ and }}\alpha \beta $ .
$\dfrac{1}{\alpha } + \dfrac{1}{\beta } = \dfrac{{\alpha + \beta }}{{\alpha \beta }}$
Also
$
\alpha + \beta = \dfrac{{ - {\text{b}}}}{{\text{a}}} = \dfrac{{ - 3}}{1} = - 3 \\
\alpha \beta = \dfrac{{\text{c}}}{a} = \dfrac{{ - 4}}{1} = - 4 \\
$
Putting the values of $\alpha + \beta {\text{ and }}\alpha \beta $ in above expression, we get:
$\dfrac{1}{\alpha } + \dfrac{1}{\beta } = \dfrac{{\alpha + \beta }}{{\alpha \beta }} = \dfrac{{ - 3}}{{ - 4}} = \dfrac{3}{4}$ .
So option B is correct.
Note: In this type of question, we should know the general expression of a quadratic equation in terms of a, b and c. We should also remember the relation between the sum and product of roots and the coefficients of the quadratic equation ${\text{a}}{{\text{x}}^2} + {\text{bx + c = 0}}$. Finally convert the given expression in the terms of $\alpha + \beta {\text{ and }}\alpha \beta $.
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