Complex Numbers And Quadratic Equations question 637
Question: The values of k for which the equations $ x^{2}-kx-21=0 $ and $ x^{2}-3kx+35=0 $ will have a common roots are:
Options:
A) $ k=\pm 4 $
B) $ k=\pm 1 $
C) $ k=\pm 3 $
D) $ k=0 $
Show Answer
Answer:
Correct Answer: A
Solution:
Let $ \alpha $ be the common root to the equations: $ x^{2}-kx-21=0 $ and $ x^{2}-3kx+35=0 $
$ \therefore ‘\alpha ’ $ satisfies both the equations
$ \therefore {{\alpha }^{2}}-k\alpha -21=0 $ …(i) and
$ \therefore {{\alpha }^{2}}-3k\alpha +35=0 $ …(ii) From (i) and (ii), $ {{\alpha }^{2}}-21=\frac{{{\alpha }^{2}}+35}{3} $
$ \Rightarrow 3{{\alpha }^{2}}-63={{\alpha }^{2}}+35 $
$ \Rightarrow {{\alpha }^{2}}=49\Rightarrow \alpha =\pm 7 $ Now, again by eliminating $ {{\alpha }^{2}} $ from (i) and (ii), we get $ k\alpha +21=3k\alpha -35 $
$ \Rightarrow 2k\alpha =56\Rightarrow k=\frac{56}{2\alpha } $ When $ \alpha =7 $ then $ k=4 $ When $ \alpha =-7 $ then $ k=-4 $ Hence, $ k=\pm 4 $
BETA
