If the equations $k (6 x^{2} + 3) + r x + 2 x^{2} - 1 = 0 a n d 6 k (2 x^{2} + 1) + p x + 4 x^{2} - 2 = 0$ have both the roots common, find the value of $2 r - p$

Open in App

Solution

The two equations can be written as

$x^{2} (6 k + 2) + r x + (3 k - 1) = 0$ ...(1)

and $x^{2} (12 k + 4) + p x + (6 k - 2) = 0.$ ...(2)

Divide by $2$

$∴ x^{2} (6 k + 2) + \frac{p}{2} x + (3 k - 1) = 0$ ...(3)

Comparing (1) and (3), we get

$r = \frac{p}{2} ∴ 2 r - p = 0.$

Suggest Corrections

Similar questions

k (6 x^{2} + 3) + r x + (2 x^{2} - 1) = 0

6 k (2 x^{2} + 1) + p x + (4 x^{2} - 2) = 0

\frac{p}{r}

k (6 x^{2} + 3) + r x + 2 x^{2} - 1 = 0

6 k (2 x^{2} + 1) + p x + 4 x^{2} - 2 = 0

2 r - p

k (6 x^{2} + 3) + r x + (2 x^{2} - 1) = 0

6 k (2 x^{2} + 1) + p x + 4 x^{2} - 2 = 0

2 r - p

k (6 x^{2} + 3) + r x + (2 x^{2} - 1) = 0

6 k (2 x^{2} + 1) + p x + (4 x^{2} - 2) = 0, k \neq 0

\frac{p}{r}

If both the roots of $k (6 x^{2} + 3) + r x + 2 x^{2} - 1 = 0$ and $2 (6 k + 2) x^{2} + p x + 2 (3 k - 1) = 0$ are common, then 2r-p is equal to

Join BYJU'S Learning Program