I need help Solving for x and finding the zeros of f(x) (please help me)
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I need help Solving for x and finding the zeros of f(x) (please help me)

[From: ] [author: ] [Date: 11-10-19] [Hit: ]
-4(x-2)(x-1+1)=0Add 1 to -1 to get 0.-4(x-2)(x)=0Remove the parentheses.-4*x(x-2)=0Multiply -4 by x to get -4x.(-4x)(x-2)=0Remove the parentheses.-4x(x-2)=0If any individual factor on the left-hand side of the equation is equal to 0, the entire expression will be equal to 0.......

The binomial can be factored using the difference of squares formula, because both terms are perfect squares. The difference of squares formula is a^(2)-b^(2)=(a-b)(a+b).
-4((x-1)-1)((x-1)+1)=0

Remove the parentheses around the expression x-1.
-4(x-1-1)((x-1)+1)=0

Subtract 1 from -1 to get -2.
-4(x-2)((x-1)+1)=0

Remove the parentheses around the expression x-1.
-4(x-2)(x-1+1)=0

Add 1 to -1 to get 0.
-4(x-2)(x)=0

Remove the parentheses.
-4*x(x-2)=0

Multiply -4 by x to get -4x.
(-4x)(x-2)=0

Remove the parentheses.
-4x(x-2)=0

If any individual factor on the left-hand side of the equation is equal to 0, the entire expression will be equal to 0.
-4x=0_(x-2)=0

Set the first factor equal to 0 and solve.
-4x=0

Divide each term in the equation by -4.
-(4x)/(-4)=(0)/(-4)

Simplify the left-hand side of the equation by canceling the common terms.
x=(0)/(-4)

Any expression with zero in the numerator is zero.
x=0

Set the next factor equal to 0 and solve.
(x-2)=0

Remove the parentheses around the expression x-2.
x-2=0

Since -2 does not contain the variable to solve for, move it to the right-hand side of the equation by adding 2 to both sides.
x=2

The final solution is all the values that make -4x(x-2)=0 true. The multiplicity of a root is the number of times the root appears. For example, a factor of has multiplicity of .
x=0,2

-
Hi,

x² - 3x = -2

x² - 3x + 2 = 0

(x - 2)(x - 1) = 0

x = 2 or x = 1 <==ANSWER



f(x) = -4(x - 1)² + 4

The vertex is at (1,4) and this graph opens down. When you one square to the left or right of the vertex, the graph drops 4 squares to have zeros at (0,0) and at (2.0). <==ANSWER

I hope that helps!! :-)

-
1. x^2 -3x = - 2

x(x - 3) = -2

1. x = -2

2. x - 3 = 0

x = 3

2. - 4(x-1)^2 + 4

-4(x^2 - 2x + 1) + 4

-4x^2 + 8x - 4 + 4

-4x(x + 2) = 0

1. -4x = 0

x = 0

2. x + 2 = 0

x = -2

-
The solutions to the first one are 1 and 2.
The solutions to the second one are 0 and 2.
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