Question

if the equation ln(x^2+2ax) =ln(4x-4a-13) has only one solution find set of values of a

Answer 1

EXPLANATION.

If the equation ln ( x² + 2ax ) = ln ( 4x - 4a - 13).

has only one solution.

To find value of a.

ln ( x² + 2ax ) = ln ( 4x - 4a - 13).

→ x² + 2ax = 4x - 4a - 13.

→ x² + 2ax - 4x + 4a + 13 = 0.

→ x² + ( 2a - 4 )x + 4a + 13 = 0.

→ it is in the form of quadratic equation

ax² + bx + c = 0.

→ D = 0 or [ b² - 4ac = 0 ].

→ ( 2a - 4 )² - 4(1)(4a + 13 ) = 0.

→ 4a² + 16 - 16a - 16a - 52 = 0.

→ 4a² - 32a - 36 = 0.

→ 4 [ a² - 8a - 9 ] = 0.

→ a² - 8a - 9 = 0.

Factories into middle term split.

→ a² - 9a + a - 9 = 0.

→ a ( a - 9 ) + 1 ( a - 9 ) = 0.

→ ( a + 1 ) ( a - 9 ) = 0.

→ a = - 1 And a = 9.

Answer 2

Answer:

This implies that

x2+2ax=4x−4a−13

or

x2+2ax−4x+4a+13=0

or

x2+(2a−4)x+(4a+13)=0

Since the equation has just one solution instead of the usual two distinct solutions, then the two solutions must be same i.e. discriminant = 0.

Hence we get that

(2a−4)2=4⋅1⋅(4a+13)

or

4a2−16a+16=16a+52

or

4a2−32a−36=0

or

a2−8a−9=0

or

(a−9)(a+1)=0

So the values of a are −1 and 9.