Question

If ${x^2} - 6x + 5 = 0$and ${x^2} - 12x + p = 0$ have a common root, then find the value of $p$.

Answer 1

Hint: If two quadratic equation ${a_1}{x^2} + {b_1}x + {c_1} = 0$ and ${a_2}{x^2} + {b_2}x + {c_2} = 0$ have one common root then the given condition ${\left( {{a_1}{c_2} - {a_2}{c_1}} \right)^2} = \left( {{a_1}{b_2} - {a_2}{b_1}} \right)\left( {{b_1}{c_2} - {b_2}{c_1}} \right)$ is satisfied by the coefficients of the given equations. So, put the coefficients of the given two equations in the above given relation and solve the equation to get the required value of $p$.

Complete step-by-step solution:
Here, the given quadratic equations are ${x^2} - 6x + 5 = 0$ and ${x^2} - 12x + p = 0$.
Comparing with the general form of a quadratic equation we get the value of different coefficient as
${a_1} = 1$, ${b_1} = - 6$, ${c_1} = 5$, ${a_2} = 1$ , ${b_2} = - 12$ and ${c_2} = p$.
We know that the condition for one common root is ${\left( {{a_1}{c_2} - {a_2}{c_1}} \right)^2} = \left( {{a_1}{b_2} - {a_2}{b_1}} \right)\left( {{b_1}{c_2} - {b_2}{c_1}} \right)$.
Now, putting the value of coefficient of the given two equations. we get,
$
   \Rightarrow {\left( {1 \times p - 1 \times 5} \right)^2} = \left( {1 \times \left( { - 12} \right) - 1 \times \left( { - 6} \right)} \right)\left( {\left( { - 6} \right) \times p - \left( { - 12} \right) \times 5} \right) \\
   \Rightarrow {\left( {p - 5} \right)^2} = \left( { - 12 + 6} \right)\left( { - 6p + 60} \right)
 $
By using a formula ${\left( {x - y} \right)^2} = {x^2} + {y^2} + 2xy$, expand ${\left( {p - 5} \right)^2}$.
$
   \Rightarrow {p^2} + 25 - 10p = \left( { - 6} \right)\left( { - 6p + 60} \right) \\
   \Rightarrow {p^2} + 25 - 10p = 36p - 360 \\
   \Rightarrow {p^2} - 46p + 385 = 0
 $
Now, we have to factorise the given equation. so, break $46p$ into two parts such that their product is $385{p^2}$.
$
   \Rightarrow {p^2} - 11p - 35p + 385 = 0 \\
   \Rightarrow p\left( {p - 11} \right) - 35\left( {p - 11} \right) = 0 \\
   \Rightarrow \left( {p - 11} \right)\left( {p - 35} \right) = 0
\ $
This implies that $\left( {p - 11} \right) = 0$ and $\left( {p - 35} \right) = 0$. So, $p = 11$ and $p = 35$.

Thus, there are two possible values of $p$ that is $11$ and $35$ for which the given two equations have one common root.

Note: Alternatively, the required value of $p$ can be calculated easily. Since the given first equation is so simple that its roots can be calculated easily. Since, the given equation has only one root common so putting the value of $x$ that is the root of the first equation in the second equation, we get the value of $p$. Since, the equation has two distinct roots so we get correspondingly two values of $p$.
If both the roots of the given equation are equal then their coefficient satisfies the relation $\dfrac{{{a_1}}}{{{a_2}}} = \dfrac{{{b_1}}}{{{b_2}}} = \dfrac{{{c_1}}}{{{c_2}}}$.