First slide

Theory of equations

Question

The quadratic equations $x^{2} - 6 x + a = 0$ and $x^{2} - c x + 6 = 0$ have one root in common. The other roots of the first and second equations are integers in the ratio 4:3. Then the common root is

Moderate

A

1
B

4
C

3
D

2

Solution

The given quadratic equations are $x^{2} - 6 x + a = 0$ …… ( 1 )
and $x^{2} - c x + 6 = 0$ ….. ( 2 )

Given that one root is common, it is consider as $α$

And the other roots of the first and second equations are integers in the ratio 4 : 3

$∴$ the roots are $α, 4 β, α, 3 β$

From Eq.( 1 ) & Eq.( 2 ) consider product of the roots

$\begin{matrix} \Rightarrow 4 α β = a a n d 3 α β = 6 \\ ∴ α β = 2, a = 8 \end{matrix}$

From ( 1 ): $x^{2} - 6 x + 8 = 0$

$\Rightarrow x^{2} - 4 x - 2 x + 8 = 0$

$\Rightarrow (x - 2) (x - 4) = 0$

$∴ x = 2, 4$     ….. ( 3 )

If take 2 is common root then Eq. ( 2 ) will be

$x^{2} - c x + 6 = 0$

$\Rightarrow {(2)}^{2} - c (2) + 6 = 0$

$\Rightarrow c = 5$ .

If take 4 is common root then Eq.( 2 )is not satisfies.

$∴$ Eq. ( 2 ) $\Rightarrow$ $x^{2} - 5 x + 6 = 0$

$\Rightarrow x^{2} - 3 x - 2 x + 6 = 0$

$\Rightarrow (x - 3) (x - 2) = 0$

$∴ x = 3, 2$     ….. ( 4 )

From Eq.( 3 ) & Eq.( 4 )

$∴$ common root =2

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