Find the value for k where the eqn x² - (k+4)x + (k-3) = 0 has one root equal to -2

Also, if roots are reciprocals of each other

and

if they are equal in absolute value, but opposite in sign.

i am totally lost on this one. thanks to anyone who can help.

Since -2 is a root, the equation should hold good when you replace x with -2.

That is,

(-2)^2 - (k+4)(-2) + (k-3) = 0
expanding,

4 + 2k + 8 + k - 3 = 0

3k + 9 = 0

k = -3

If it has a root equal to -2, then substituting -2 for x gives a zero. Thus,

(-2)^2 - (k + 4)(-2) + (k - 3) = 0
4 + 2k + 8 + k - 3 = 0
3k + 9 = 0
3k = -9
k = -3
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Let r be one root of the equation. Then, 1/r is also a root.

The product of the roots r(1/r) is clearly 1, which is also equal to k - 3.
Thus, k = 4.
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Let r be one root of the equation. Then -r is also a root.

The sum of the roots -r + r is clearly 0, which is also equal to k + 4.
Thus, k = -4.

Why all the thumbs down? Rao had a good solution, and it is right.

The problem said has one root equal to x = -2, not two repeated roots of x = -2 (which would allow equating the left hand side to (x+2)^2).

k = -3, good job, Rao.

x² - (k+4)x + (k-3) = (x+2)² = x² + 4x + 4

=> - (k + 4) = 4

=> k = -8

but -8 - 3 not equal 4 ,hence no such k

EDIT : u r right guys i thought it meant just one root equal -2 k =-3 is the righ answer :)