
If \[\alpha \] and $\beta $ are the roots of $4{x^2} + 3x + 7 = 0$ then the value of $\dfrac{1}{\alpha } + \dfrac{1}{\beta }$ is
A $\dfrac{4}{7}$
B $\dfrac{{ - 3}}{7}$
C $\dfrac{3}{7}$
D $ - \dfrac{3}{4}$
Answer
161.4k+ views
Hint:First, we will find the sum of roots and the product of roots of a given quadratic equation using the formula. Then we will $\dfrac{1}{\alpha } + \dfrac{1}{\beta }$ solve this by taking LCM and then after solving we will put the value of sum and product of roots. Then after solving we will get the required answer.
Formula Used:
The general form of the quadratic equation: $ax^2+bx+c=0$
Sum of roots $ = \dfrac{{ - b}}{a}$
And product of roots $\dfrac{c}{a}$
Complete step by step Solution:
Given, equations is $4{x^2} + 3x + 7 = 0$
Let \[\alpha \] and $\beta $ are the roots of given equation.
The general quadratic equation $a{x^2} + bx + c = 0$
We know that sum of roots $ = \dfrac{{ - b}}{a}$
And product of roots $\dfrac{c}{a}$
Using this formula
Sum of roots of $4{x^2} + 3x + 7 = 0$ is
$\alpha + \beta = \dfrac{{ - 3}}{4}$
Products of roots of $4{x^2} + 3x + 7 = 0$ is
$\alpha \beta = \dfrac{7}{4}$
$\dfrac{1}{\alpha } + \dfrac{1}{\beta } = \dfrac{{\alpha + \beta }}{{\alpha \beta }}$
Putting value of $\alpha + \beta $ and $\alpha \beta $
$ = \dfrac{{\dfrac{{ - 3}}{4}}}{{\dfrac{7}{4}}}$
$ = \dfrac{{ - 3}}{4} \times \dfrac{4}{7}$
After solving the above expressions, we will get
$ = \dfrac{{ - 3}}{7}$
Hence, the value $\dfrac{1}{\alpha } + \dfrac{1}{\beta }$ of is $\dfrac{{ - 3}}{7}$
Hence, the correct option is (B).
Additional Information:The values of the variable that satisfy a quadratic equation are known as its roots. The "solutions" or "zeroes" of the quadratic equation are other names for them. The general quadratic equation is $a{x^2} + bx + c = 0$
The roots of the equation can be determined by the following method:
1. Factorization
2. Graphing
3. Completing the square
4. Quadratic formula
And the sum of roots $ = \dfrac{{ - b}}{a}$
And product of roots $\dfrac{c}{a}$
Note: Students should solve question by taking LCM of $\dfrac{1}{\alpha } + \dfrac{1}{\beta }$ using this they will get solution easily. If they try first to find the value of \[\alpha \] or $\beta $ then putting in the product of the equation in that calculation will be more complicated and they may make mistakes.
Formula Used:
The general form of the quadratic equation: $ax^2+bx+c=0$
Sum of roots $ = \dfrac{{ - b}}{a}$
And product of roots $\dfrac{c}{a}$
Complete step by step Solution:
Given, equations is $4{x^2} + 3x + 7 = 0$
Let \[\alpha \] and $\beta $ are the roots of given equation.
The general quadratic equation $a{x^2} + bx + c = 0$
We know that sum of roots $ = \dfrac{{ - b}}{a}$
And product of roots $\dfrac{c}{a}$
Using this formula
Sum of roots of $4{x^2} + 3x + 7 = 0$ is
$\alpha + \beta = \dfrac{{ - 3}}{4}$
Products of roots of $4{x^2} + 3x + 7 = 0$ is
$\alpha \beta = \dfrac{7}{4}$
$\dfrac{1}{\alpha } + \dfrac{1}{\beta } = \dfrac{{\alpha + \beta }}{{\alpha \beta }}$
Putting value of $\alpha + \beta $ and $\alpha \beta $
$ = \dfrac{{\dfrac{{ - 3}}{4}}}{{\dfrac{7}{4}}}$
$ = \dfrac{{ - 3}}{4} \times \dfrac{4}{7}$
After solving the above expressions, we will get
$ = \dfrac{{ - 3}}{7}$
Hence, the value $\dfrac{1}{\alpha } + \dfrac{1}{\beta }$ of is $\dfrac{{ - 3}}{7}$
Hence, the correct option is (B).
Additional Information:The values of the variable that satisfy a quadratic equation are known as its roots. The "solutions" or "zeroes" of the quadratic equation are other names for them. The general quadratic equation is $a{x^2} + bx + c = 0$
The roots of the equation can be determined by the following method:
1. Factorization
2. Graphing
3. Completing the square
4. Quadratic formula
And the sum of roots $ = \dfrac{{ - b}}{a}$
And product of roots $\dfrac{c}{a}$
Note: Students should solve question by taking LCM of $\dfrac{1}{\alpha } + \dfrac{1}{\beta }$ using this they will get solution easily. If they try first to find the value of \[\alpha \] or $\beta $ then putting in the product of the equation in that calculation will be more complicated and they may make mistakes.
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