Determine the value(s) of k such that the equation x^2 +3x-k has:

Question

1. Two distinct roots, 2 one unique real root, 3 no real root?

Answer 1

(1)

Given polynomial : f(x) = x² +  3x - k

To find roots of this polynomial

Compare f(x) = x² +  3x - k with general quadratic form : x² + bx + c

a = 1 , b = 3 , c = - k

Discriminant D = ( b² - 4ac )

Substitute : a = 1 , b = 3 , c = - k

D = ( 3² - 4(1)(-k) )

D = ( 9 + 4k )

f(x) = x² +  3x - k has two distinct real roots when D > 0

D > 0

( 9 + 4k ) > 0

4k  > -9

k  > - 9/4

Answer 2

(3)

Given polynomial : f(x) = x² +  3x - k

To find roots of this polynomial

Compare f(x) = x² +  3x - k with general quadratic form : x² + bx + c

a = 1 , b = 3 , c = - k

Discriminant D = ( b² - 4ac )

Substitute : a = 1 , b = 3 , c = - k

D = ( 3² - 4(1)(-k) )

D = ( 9 + 4k )

f(x) = x² +  3x - k has two Non real roots ( imaginary/complex roots) when D < 0

D < 0

( 9 + 4k ) < 0

4k  < - 9

k  < - 9/4

Answer 3

(2)

Given polynomial : f(x) = x² +  3x - k

To find roots of this polynomial

Compare f(x) = x² +  3x - k with general quadratic form : x² + bx + c

a = 1 , b = 3 , c = - k

Discriminant D = ( b² - 4ac )

Substitute : a = 1 , b = 3 , c = - k

D = ( 3² - 4(1)(-k) )

D = ( 9 + 4k )

f(x) = x² +  3x - k has only one real root ( equal roots) when D = 0

D = 0

( 9 + 4k ) = 0

4k  = - 9

k  = - 9/4