Question

For what value of k is there one solution to the given quadratic equation (k+1)x²+4kx+2=0.

Answer 1

If
`k = 1` or
`k = -(1)/(2)`, there is only one solution to the given quadratic equation.

Explanation:

Given a second order polynomial expressed by the following equation:

`ax^(2) + bx + c, a\\eq0`

This polynomial has roots
`x_(1), x_(2)` such that
`ax^(2) + bx + c = (x - x_(1))*(x - x_(2))`, given by the following formulas:

`x_(1) = (-b + √(\bigtriangleup))/(2*a)`

`x_(2) = (-b - √(\bigtriangleup))/(2*a)`

`\bigtriangleup = b^(2) - 4ac`

The signal of
`\bigtriangleup` determines how many real roots an equation has:

`\bigtriangleup > 0`: Two real and different solutions

`\bigtriangleup = 0`: One real solution

`\bigtriangleup < 0`: No real solutions

In this problem, we have the following second order polynomial:

`(k+1)x^(2) + 4kx + 2 = 0`.

This means that
`a = k+1; b = 4k; c = 2`

It has one solution if

`\bigtriangleup = 0`

`b^(2) - 4ac = 0`

`16k^(2) -8(k+1) = 0`

`16k^(2) - 8k - 8 = 0`

We can simplify by 8

`2k^(2) - k - 1 = 0`

The solution is:

`k = 1` or
`k = -(1)/(2)`

So, if
`k = 1` or
`k = -(1)/(2)`, there is only one solution to the given quadratic equation.