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# (k - 2) x^2 + (k - 5) x - 5 = 0?

k =/= 2,

Find k if the roots are equal...

### 3 Answers

- PuggyLv 71 decade agoFavorite Answer
(k - 2)x^2 + (k - 5)x - 5 = 0

The roots are equal if the discriminant is equal to zero.

The discriminant, b^2 - 4ac, evaluated as follows:

(k - 5)^2 - 4(k - 2)(-5)

And we want this to be equal to zero.

(k - 5)^2 - 4(k - 2)(-5) = 0

(k - 5)^2 + 20(k - 2) = 0

k^2 - 10k + 25 + 20k - 40 = 0

k^2 + 10k - 15 = 0

k = [ -10 +/- sqrt(100 - 4(-15)) ] / 2

k = [ -10 +/- sqrt(100 + 60) ] / 2

k = [ -10 +/- sqrt(160) ] / 2

k = [ -10 +/- sqrt(16*10) ] / 2

k = [ -10 +/- 4sqrt(10) ] / 2

k = -5 +/- 2sqrt(10)

- Moise GunenLv 71 decade ago
Delta=0 if roots are equal

then

(k-5)^2+20(k-2)=0

k^2-10k+25+20k-40=0

K^2+10k-15=0

k=5+/-sqrt(20)=5+/-2sqrt(5)

- 1 decade ago
if roots are equal

deta=b^2-4ac=0

b^2-4ac=(k-5)^2-4*(k-2)*(-5)

=k^2+10k-15=0

k^2+10k-15=0

k^2+10k+25=40

(k+5)^2=40

k=sqt40-5or-sqt40-5

k=2sqt10-5 or -2sqt10-5