
If two equations ${x^2} - cx + d = 0$ and ${x^2} - ax + b = 0$ have one common root and the second has equal roots, then $2(b + d) = $
A. $0$
B. $a + c$
C. $ac$
D. $ - ac$
Answer
163.2k+ views
Hint: We have to establish a relation between coefficients of the given quadratic equations. To proceed further we have to find the sum and product of the roots of the given quadratic equations. For the first equation we will simply find the sum and product of the roots. But according to the question, the second equation has equal roots. Taking this under consideration, we will find the sum and product of the equal roots too. And at last, substitute all these values accordingly to get the result.
Formula Used: For a quadratic equation $a{x^2} + bx + c = 0$ having roots $\alpha ,\beta $ . The sum of roots is: $(\alpha + \beta ) = - \dfrac{b}{a}$ and the product of roots is: $(\alpha \beta ) = \dfrac{c}{a}$ .
Complete step-by-step solution:
We have two quadratic equations
${x^2} - cx + d = 0$ ----(1.1)
${x^2} - ax + b = 0$ ----(1.2)
If we compare equations (1.1) and (1.2) with ${a_1}{x^2} + {b_1}x + {c_1} = 0$ and ${a_2}{x^2} + {b_2}x + {c_2} = 0$, we will get ${a_1} = 1$ , ${b_1} = - c$ , ${c_1} = d$ ; ${a_2} = 1$ , ${b_2} = - a$ , ${c_2} = b$ .
Suppose, the given equation (1.1) has roots $\alpha ,\beta $ .
Then, the sum of roots will be:
$(\alpha + \beta ) = - \dfrac{{{b_1}}}{{{a_1}}}$
$ \Rightarrow (\alpha + \beta ) = c$
And the product of roots:
$(\alpha \beta ) = \dfrac{{{c_1}}}{{{a_1}}}$
$ \Rightarrow (\alpha \beta ) = d$
Similarly, equation (1.2) has the equal roots, say, $\alpha ,\alpha $ .
$\therefore $ The sum of roots
$ \Rightarrow (\alpha + \alpha ) = 2\alpha = a$
And the product of the roots will be
$ \Rightarrow (\alpha \cdot \alpha ) = {\alpha ^2} = b$
Now, to find the value of $2(b + d)$ , substitute the values of $b$ and $d$ into it, we get
$ \Rightarrow 2(b + d) = 2({\alpha ^2} + \alpha \beta )$
$ \Rightarrow 2(b + d) = 2\alpha (\alpha + \beta )$
We have $2\alpha = a$ and $(\alpha + \beta ) = c$
So,
$ \Rightarrow 2(b + d) = ac$
Hence, the correct option is C.
Note: As it is given in the question that the equations have one common root. So, we can also use the method of cramer’s rule to get the value of the common root. And then, we can proceed with the process of finding the sum of the roots and product of the roots. And the second equation has equal roots, so the process for this will be the same.
Formula Used: For a quadratic equation $a{x^2} + bx + c = 0$ having roots $\alpha ,\beta $ . The sum of roots is: $(\alpha + \beta ) = - \dfrac{b}{a}$ and the product of roots is: $(\alpha \beta ) = \dfrac{c}{a}$ .
Complete step-by-step solution:
We have two quadratic equations
${x^2} - cx + d = 0$ ----(1.1)
${x^2} - ax + b = 0$ ----(1.2)
If we compare equations (1.1) and (1.2) with ${a_1}{x^2} + {b_1}x + {c_1} = 0$ and ${a_2}{x^2} + {b_2}x + {c_2} = 0$, we will get ${a_1} = 1$ , ${b_1} = - c$ , ${c_1} = d$ ; ${a_2} = 1$ , ${b_2} = - a$ , ${c_2} = b$ .
Suppose, the given equation (1.1) has roots $\alpha ,\beta $ .
Then, the sum of roots will be:
$(\alpha + \beta ) = - \dfrac{{{b_1}}}{{{a_1}}}$
$ \Rightarrow (\alpha + \beta ) = c$
And the product of roots:
$(\alpha \beta ) = \dfrac{{{c_1}}}{{{a_1}}}$
$ \Rightarrow (\alpha \beta ) = d$
Similarly, equation (1.2) has the equal roots, say, $\alpha ,\alpha $ .
$\therefore $ The sum of roots
$ \Rightarrow (\alpha + \alpha ) = 2\alpha = a$
And the product of the roots will be
$ \Rightarrow (\alpha \cdot \alpha ) = {\alpha ^2} = b$
Now, to find the value of $2(b + d)$ , substitute the values of $b$ and $d$ into it, we get
$ \Rightarrow 2(b + d) = 2({\alpha ^2} + \alpha \beta )$
$ \Rightarrow 2(b + d) = 2\alpha (\alpha + \beta )$
We have $2\alpha = a$ and $(\alpha + \beta ) = c$
So,
$ \Rightarrow 2(b + d) = ac$
Hence, the correct option is C.
Note: As it is given in the question that the equations have one common root. So, we can also use the method of cramer’s rule to get the value of the common root. And then, we can proceed with the process of finding the sum of the roots and product of the roots. And the second equation has equal roots, so the process for this will be the same.
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