Quadratic Equations and Expressions
4.0 Common roots
4.0 Common roots
Consider two quadratic equations $${a_1}{x^2} + {b_1}x + {c_1} = 0\quad ...(1)$$ and $${a_2}{x^2} + {b_2}x + {c_2} = 0\quad ...(2)$$
where ${a_1}{a_2} \ne 0$ and ${a_1}{b_2} - {a_2}{b_1} \ne 0$
If the equations $(1)$ and $(2)$ have one root in common, say $\alpha $, then it must satisfy both the equations i.e., $${a_1}{\alpha ^2} + {b_1}\alpha + {c_1} = 0\quad ...(3)$$ and $${a_2}{\alpha ^2} + {b_2}\alpha + {c_2} = 0\quad ...(4)$$
Solving $(3)$ and $(4)$, we get $$\frac{{{\alpha ^2}}}{{{b_1}{c_2} - {b_2}{c_1}}} = \frac{\alpha }{{{c_1}{a_2} - {c_2}{a_1}}} = \frac{1}{{{a_1}{b_2} - {a_2}{b_1}}}$$
On solving, we get $$\begin{equation} \begin{aligned} \alpha = \frac{{{b_1}{c_2} - {b_2}{c_1}}}{{{c_1}{a_2} - {c_2}{a_1}}} = \frac{{{c_1}{a_2} - {c_2}{a_1}}}{{{a_1}{b_2} - {a_2}{b_1}}} \\ \Rightarrow {({c_1}{a_2} - {c_2}{a_1})^2} = \left( {{a_1}{b_2} - {a_2}{b_1}} \right)\left( {{b_1}{c_2} - {b_2}{c_1}} \right) \\\end{aligned} \end{equation} $$
If the equations $(1)$ and $(2)$ have both roots common, then the equations are identical and the co-efficients are in proportion i.e., $$\frac{{{a_1}}}{{{a_2}}} = \frac{{{b_1}}}{{{b_2}}} = \frac{{{c_1}}}{{{c_2}}}$$
Note:
- To find the common root between the two equations, make the co-efficient of ${x^2}$ same in both the equations and then subtract them.
- If $f(x) = 0$ and $g(x) = 0$ are two polynomial equation having some common root(s) then those common root(s) is/are also the root(s) of $h(x)=af(x)+bg(x)=0$.
Question 4. Find the value of $k$, so that the equations ${x^2} - x - 12 = 0$ and $k{x^2} + 10x + 3 = 0$ may have one root in common. Also find the common root.
Solution: Let $\alpha $ be the common root of the two equations. Hence, ${\alpha ^2} - \alpha - 12 = 0$ and $k{\alpha ^2} + 10\alpha + 3 = 0$. Solving them and eliminating $\alpha $, we get $$\begin{equation} \begin{aligned} {( - 12k - 3)^2} = 117(10 + k) \\ {(4k + 1)^2} = 13(10 + k) \\ 16{k^2} - 5k - 129 = 0 \\ 16{k^2} - 48k + 43k - 129 = 0 \\ k = 3{\text{ or }}k = - \frac{{43}}{{16}} \\\end{aligned} \end{equation} $$
Using value of $k$, $$\alpha = \frac{{ - 12k - 3}}{{10 + k}} = - 3{\text{ or }}4$$
