Question

I know the answer but don't know the full solution. Please help​

Answer 1

Solution :-

Equations :-

1.) 2k(x² + 2) + rx + x² - 5 = 0

2.) 2k( 3x² + 1) + px - x² - 1 = 0

Both of them have common roots.

Now by expansion of the equations.

1.) 2k(x² + 2) + rx + x² - 5 = 0

→ 2kx² + 4k + rx + x² - 5 = 0

→ x²(2k + 1) + rx + (4k - 5) = 0

2.) 2k( 3x² + 1) + px - x² - 1 = 0

→ 6kx² + 2k + px - x² - 1 = 0

→ x²(6k - 1) + px + (2k - 1) = 0

Now applying the condition for the common roots.

$\dfrac{a}{a'} = \dfrac{b}{b'} = \dfrac{c}{c'}$

$\dfrac{(2k + 1)}{(6k-1)} = \dfrac{r}{p} = \dfrac{(4k-5)}{(2k -1)}$

Now by taking

$\dfrac{(2k + 1)}{(6k-1)} \: and \:\dfrac{(4k-5)}{(2k -1)}$

$\implies \dfrac{(2k + 1)}{(6k-1)} = \dfrac{(4k-5)}{(2k -1)}$

By cross multiplication :-

$\implies (2k + 1)(2k-1) = (4k - 5)(6k -1)$

$\implies 4k^2 - 1 = 24k^2 - 4k - 30k + 5$

$\implies 20k^2 - 34k + 6 = 0$

$\implies 10k^2 - 17k + 3 = 0$

$\implies 10k^2 - 15k - 2k + 3 = 0$

$\implies 5k( 2k - 3) -1(2k - 3) = 0$

$\implies (2k -3)(5k -1) = 0$

$\implies k = \dfrac{1}{5} \: or \: \dfrac{3}{2}$

Now as k > 1

→ k = 3/2

Now

$\implies \dfrac{r}{p} = \dfrac{(4k-5)}{(2k -1)}$

$\implies \dfrac{r}{p} = \dfrac{(4(3/2) -5)}{(2(3/2) -1)}$

$\implies \dfrac{r}{p} = \dfrac{(6 - 5)}{(3 -1)}$

$\implies \dfrac{r}{p} = \dfrac{1}{2}$

$\implies 2r = p$

Then

= 2r - p

= 2r - 2r

= 0