Question

If x2 -3ax+14 = 0 and x2 + 2ax-16
= 0have a common root, then the value of a is​

Answer 1

The value of a comes out to be ± 3.

Given,

x² - 3ax + 14 = 0 and x² + 2ax - 16 = 0 have a common root

To Find,

Value of 'a'

Solution,

Let the common root for both the equations be 'k'

Therefore,

x² - 3ax + 14 = 0

k² - 3ak + 14 = 0

a = $\frac{k^2+14}{3k}$

Substituting the value of 'a' in the second equation

x² + 2ax - 16 = 0

k² + 2ak - 16 = 0

k² + 2 * ($\frac{k^2+14}{3k}$) * k - 16 = 0

k² + 2 * ($\frac{k^2+14}{3}$) - 16 = 0

3k² + 2k² + 28 - 48 = 0

5k² = 20

k² = 4

k = ± 2

Now, a = $\frac{k^2+14}{3k}$

If k = + 2

a = $\frac{4+14}{2*3} = \frac{18}{6} = 3$

If k = - 2

a = $\frac{4+14}{-2*3} = \frac{18}{-6} = -3$

Thus, the value of a comes out to be ± 3.

#SPJ2