Quadratic Equations by Factoring 2x(x+3)=0 and 2x^2+9x-5=0 Can anyone help me?

Question

Quadratic Equations by Factoring 2x(x+3)=0 and 2x^2+9x-5=0 Can anyone help me?

Answer 1

2x(x + 3) = 0 (zero-product property)

2x = 0 or x + 3 = 0

x = 0 or x = -3 <===ANSWER

2x^2 + 9x - 5 = 0 (factor)

(2x - 1)(x + 5) = 0 (zero-product property)

2x - 1 = 0 or x + 5 = 0

2x = 1 or x = -5

x = 1/2 or x = -5 <===ANSWER

Answer 2

The first one is factored and the answer is x=0 or x = -3

2nd is (x+5)(2x-1) so x= -5 or x=1/2

Answer 3

2x(x+3) = 0 Zero product property

2x = 0 Divide

x = 0

x+3 = 0 Subtract 3 from both sides

x = -3

2x^2 + 9x - 5 = 0 Factor using AC method

2x^2 + 10x - x - 5 = 0 Factor 2x^2 + 10x and -x - 5

2x(x+5) - 1(x+5) Pull out x+5

(x+5) What is left when you pull out the x+5? 2x-1

(2x-1)(x+5) = 0 Zero Product Property

2x - 1 = 0

x = 1/2

x + 5 = 0

x = -5

Plug all the answers into the original equation and check for any extraneous solutions.

Answer 4

Question Number 1 :

Question Number 1 :

For this equation ( 2*x  ) * ( x + 3 ) = 0 , answer the following questions :

A. Use factorization to find the root of the equation !



Answer Number 1 :

First, we must turn this equation ( 2*x  ) * ( x + 3 ) = 0 into a*x^2+b*x+c=0 form.

( 2*x  ) * ( x + 3 ) = 0 , expand the left hand side.

<=> 2*x  * ( x + 3 )   +   0 * ( x + 3 ) = 0

<=> 2*x^2 + 6*x  = 0 , move 0 from the right hand side to the left hand side.

<=> 2*x^2 + 6*x  - 0 = 0

<=> 2*x^2 + 6*x  = 0

The equation 2*x^2 + 6*x  = 0 is already in a*x^2+b*x+c=0 form.

So we can imply that the value of a = 2, b = 6, c = 0.

1A. Use factorization to find the root of the equation !

  2*x^2 + 6*x  = 0

  <=> 2 * ( x  ) * ( x + 3 ) = 0

  So we have the answers x1 = 0 and x2 = -3

Question Number 2 :

For this equation 2*x^2 + 9*x - 5 = 0 , answer the following questions :

A. Use factorization to find the root of the equation !

Answer Number 2 :

The equation 2*x^2 + 9*x - 5 = 0 is already in a*x^2+b*x+c=0 form.

By matching the constant position, we can derive that the value of a = 2, b = 9, c = -5.

2A. Use factorization to find the root of the equation !

  2*x^2 + 9*x - 5 = 0

  <=> ( 2x - 1 ) * ( x + 5 ) = 0

  So we have the answers x1 = 0.5 and x2 = -5

Answer 5

the first one is already factored:

2x(x+3) = 0

meaning 2x = 0 or x+3 =0

so x=0 or -3

2x² +9x -5 =0

(2x-1)(x+5)= 0

2x-1 =0 or x+5=0

x=1/2 or -5

Answer 6

this first is factored as far as it can go.

the second is (2x-1)(x+5)