
If \[\alpha \] and $\beta $ are the roots of the equation \[{x^2} - \;6x\; + \;a\; = \;0\;\] and satisfy the relation \[3\alpha \; + \;2\beta \; = \;16\], then the value of $a$ is
A. $ - 8$
B. $8$
C. $16$
D. $9$
E. None of these
Answer
163.2k+ views
Hint: In this question, we are given that \[\alpha \] and $\beta $ are the roots of the equation \[{x^2} - \;6x\; + \;a\; = \;0\;\]. Also, the relation between the roots is \[3\alpha \; + \;2\beta \; = \;16\]. We have to calculate the value of $a$. Calculate the sum and product of the root i.e., $\alpha + \beta = \dfrac{{ - B}}{A}$, $\alpha \beta = \dfrac{C}{A}$ where the equation is \[A{x^2} + Bx + C = 0\]. Put the equation of sum in the relation given to calculate the value of \[\alpha \] and $\beta $. In last put the required values in the product.
Formula Used:
General quadratic equation: – \[A{x^2} + Bx + C = 0\]
Let, the roots of the above quadratic equation be $\alpha $ and $\beta $
Therefore,
Sum of roots, $\alpha + \beta = \dfrac{{ - B}}{A}$
Product of roots, $\alpha \beta = \dfrac{C}{A}$
Complete step by step Solution:
Given that,
\[\alpha \] and $\beta $ are the roots of the equation \[{x^2} - \;6x\; + \;a\; = \;0\;\]
And the relation between the roots is \[3\alpha \; + \;2\beta \; = \;16\] ------(1)
Compare the equation \[{x^2} - \;6x\; + \;a\; = \;0\;\] with the general quadratic equation \[A{x^2} + Bx + C = 0\]
We get, $A = 1,B = - 6,C = a$
Now, using the formula of sum and product of the root
Sum of the roots, \[\alpha + \beta = \dfrac{{ - B}}{A} = 6\]
It implies that, \[\alpha = 6 - \beta \] --------(2)
And the product of the roots, $\alpha \beta = \dfrac{C}{A}$
$ \Rightarrow \alpha \beta = a$ --------(3)
Put \[\alpha = 6 - \beta \] in equation (1)
We get, \[3\left( {6 - \beta } \right) + \;2\beta \; = \;16\]
$18 - 3\beta + 2\beta = 16$
On solving, \[\beta = 2\]
Substitute \[\beta = 2\] in equation (2),
We get $\alpha = 4$
Now, to calculate the value of $a$ put $\alpha = 4$ and \[\beta = 2\] in equation (3),
It will be, $a = 8$
Thus, the value of $a$ is $8$.
Hence, the correct option is (B).
Note: The values of $x$ that fulfil a given quadratic equation \[A{x^2} + Bx + C = 0\] are known as its roots. They are, in other words, the values of the variable $\left( x \right)$ that satisfy the equation. The roots of a quadratic function are the $x - $ coordinates of the function's $x - $ intercepts. Because the degree of a quadratic equation is $2$, it can only have two roots.
Formula Used:
General quadratic equation: – \[A{x^2} + Bx + C = 0\]
Let, the roots of the above quadratic equation be $\alpha $ and $\beta $
Therefore,
Sum of roots, $\alpha + \beta = \dfrac{{ - B}}{A}$
Product of roots, $\alpha \beta = \dfrac{C}{A}$
Complete step by step Solution:
Given that,
\[\alpha \] and $\beta $ are the roots of the equation \[{x^2} - \;6x\; + \;a\; = \;0\;\]
And the relation between the roots is \[3\alpha \; + \;2\beta \; = \;16\] ------(1)
Compare the equation \[{x^2} - \;6x\; + \;a\; = \;0\;\] with the general quadratic equation \[A{x^2} + Bx + C = 0\]
We get, $A = 1,B = - 6,C = a$
Now, using the formula of sum and product of the root
Sum of the roots, \[\alpha + \beta = \dfrac{{ - B}}{A} = 6\]
It implies that, \[\alpha = 6 - \beta \] --------(2)
And the product of the roots, $\alpha \beta = \dfrac{C}{A}$
$ \Rightarrow \alpha \beta = a$ --------(3)
Put \[\alpha = 6 - \beta \] in equation (1)
We get, \[3\left( {6 - \beta } \right) + \;2\beta \; = \;16\]
$18 - 3\beta + 2\beta = 16$
On solving, \[\beta = 2\]
Substitute \[\beta = 2\] in equation (2),
We get $\alpha = 4$
Now, to calculate the value of $a$ put $\alpha = 4$ and \[\beta = 2\] in equation (3),
It will be, $a = 8$
Thus, the value of $a$ is $8$.
Hence, the correct option is (B).
Note: The values of $x$ that fulfil a given quadratic equation \[A{x^2} + Bx + C = 0\] are known as its roots. They are, in other words, the values of the variable $\left( x \right)$ that satisfy the equation. The roots of a quadratic function are the $x - $ coordinates of the function's $x - $ intercepts. Because the degree of a quadratic equation is $2$, it can only have two roots.
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