Algebra help!!!!!!!!!!!!!!!!!!!!!!!!!!…

Question

Find the zeros of each function

1. f(x) = x^2 = 4x - 5

2. g(x) = -x^2 + 6x - 8

3. f(x) = x^2 - 1

Please explain how you found the answer, thanks in advance!

Answer 1

1. The function  is   f(x)

 f(x) = x² - 4x - 5

Zero of the function f(x) is f(0)

Substitute the x=0 in the function f(x) =x² -4x - 5

Therefore f(0) = 0² - 4(0) - 5

f(0) = 0 - 0 - 5

Therefore f(0) = -5

Zero of the function f(x) is  -5.

Comments

Zeros of f(x) = x² - 4x - 5 are 5, -1.

Answer 2

2. The function  is   g(x)

 g(x) = -x² + 6x - 8

Zero of the function g(x) is g(0)

Substitute the x=0 in the function g(x) = - x² + 6x - 8

Therefore g(0) = - 0² + 6(0) - 8

f(0) = 0 + 0 - 8

Therefore f(0) = -8

Zero of the function g(x) is  -8.

Comments

Zeros of g (x ) = - x ² + 6x - 8 are 4, 2.

Answer 3

3.     The function  is   f(x)

 f(x) = x² - 1

Zero of the function f(x) is f(0)

Substitute the x=0 in the function f(x) = x² - 1

Therefore f(0) = 0 - 1

f(0) = - 1

Zero of the function f(x) is  -1.

 

Comments

Zeros of f(x) = x² - 1 are 1, -1.

Answer 4

1)  f(x) = x² - 4x - 5

The zeros of a function f are found by solving the equation f(x) = 0

x² - 4x - 5 = 0

Now solve  the equation x² - 4x - 5 = 0 in factorisation method.

Multiply first term x² and last term - 5 = - 5x²

The correct pair of the terms - 5x and x multiply to - 5x² and add to - 4x.

Replace the middle term - 4x with - 5x + x.

 x² - 5x + x - 5 = 0

Group the terms into two pairs.

 (x² - 5x) + (x - 5) = 0

Factor out x from the first group  and factor out 1 from the second group.

 x(x - 5) + 1(x - 5) = 0

Factor out common term x - 5.

 (x - 5)(x + 1) = 0

Apply zero product property.

x - 5 = 0 and x + 1 = 0

x = 5 and x = -1

Zeros of f(x) = x² - 4x - 5 are 5, -1.

 

Answer 5

2) The function g (x ) = - x² + 6x - 8

The zeros of a function g are found by solving the equation g (x ) = 0

- x² + 6x - 8 = 0

Now solve the equation - x² + 6x - 8 = 0 by using quadratic formula.

Compare it to quadratic form

a = - 1, b = 6 , c = -8

b² - 4ac  = (6)² - 4(-1)(-8) = 36 - 32 = 4

b² - 4ac > 0

The discriminant is positive ,The equation has two real roots.

Roots are

Zeros of g (x ) = - x ² + 6x - 8 are 4, 2.

Answer 6

The finction f(x) = x² - 1

The zeros of a function f are found by solving the equation f(x) = 0

x² - 1 = 0

x² - 1² = 0

Difference of two squares  a ²- b ² = (a + b ) (a - b )

In this case a = x , b = 1

(x - 1) (x + 1) = 0

Apply zero product property.

x - 1 = 0 and x + 1 = 0

x = 1 and x = -1

Zeros of f(x) = x² - 1 are 1, -1.