Question

consider the function f(x) = x^2 + 6x + 8 how many and what type of roots does the equation f(x) = 0 have

Answer 1

Answer:

two roots -2 and -4

Step-by-step explanation:

since

x^2+6x+8=0

x^2+2x+4x+8=0

x(x+2)+4(x+2)=0

(x+2)(x+4)=0

either x+2=0 which give us x= -2

or x+4=0 which give us x= -4 hence given quadratic equation has two real roots which are -2 and -4

Answer 2

Answer:

Step-by-step explanation:

Given that,

f ( x ) = x² + 6x + 8 = 0

⇒x² + 2x + 4x + 8 = 0

⇒ x ( x + 2 ) + 4 ( x + 2 ) = 0

⇒ ( x + 2 ) ( x + 4 ) = 0

⇒ x = - 2 and - 4.

∴There are two rooroots have f ( x ).

Let the given equation comparing with ax² + bx + c = 0,

We have a = 1 , b = 6 and c = 8.

∴Discriminant = d = b² - 4ac = ( 6 )² - 4 ( 1 ) ( 8 ) = 36 - 32 = 4.

so, d > 0.

Hence , f ( x ) have two distinct real roots.