Find k

[greatwhiteshark]

Find k such that the equation k(x^2) + x + k = 0 has a repeated real solution.

 

[Denis]

Find k such that the equation k(x^2) + x + k = 0 has a repeated real solution.

using quadratic:
x = [-1 +- sqrt(1 - 4k^2)] / 2k

if k = 1/2, then:
x = [-1 +- sqrt(1 - 1)] / 1
x = -1

I don't know what "repeated" real solution means...

 

[Gene]

Denis has it.
.5x²+x+.5 =
.5*(x²+2x+1) =
.5*(x+1)² = 0
x= -1,-1 repeated
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Gene

 

[pka]

Actually there are two answers: k=(1/2) and (−1/2)

If the discriminant, b<SUP>2</SUP>−4ac, equals 0 then ax<SUP>2</SUP>+bx+c=0 has a double root.

 

[greatwhiteshark]

ok

Thank you all.