Solve the quadratic equation by factoring.
5x2 - 8 = 39x

5x^2 - 8 = 39x; 5x^2 - 39x - 8 = 0; (5x + 1)(x - 8) = 0; 5x + 1 = 0; 5x = -1; x = -1/5; x - 8 = 0; x = 8 x = -1/5 or x = 8

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Asked 12/4/2012 12:57:30 PM

Updated 16 days ago|8/30/2014 6:22:45 PM

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Flagged by andrewpallarca [8/30/2014 6:22:38 PM]

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5x^2 - 8 = 39x;

5x^2 - 39x - 8 = 0;

(5x + 1)(x - 8) = 0;

5x + 1 = 0; 5x = -1; x = -1/5;

x - 8 = 0; x = 8

x = -1/5 or x = 8

5x^2 - 39x - 8 = 0;

(5x + 1)(x - 8) = 0;

5x + 1 = 0; 5x = -1; x = -1/5;

x - 8 = 0; x = 8

x = -1/5 or x = 8

Added 16 days ago|8/30/2014 6:22:45 PM

This answer has been confirmed as correct, not copied, and helpful.

apply the quadratic formula to find the roots of the given function, and then graph the function:f(x) = x2 + 4

Weegy: f(x)=x^2-4
a=1,b=0,c=-4
x1= (0+sqrt(0+16))/2
x1=2
x2= (0-sqrt(0+16))/2
x2= -2 (More)

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Expert Answered

Asked 11/27/2012 11:26:06 AM

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complete the square, and find the roots of the quadratic equation.x2 - 2/3x = 0 x = 1/2 (3-sqrt(37)) x = 1/2 (3+sqrt(37))

Weegy: x^2-(2/3)*x=0
x(x-(2/3)) = 0
so x=0 or x-(2/3)=0
- > x=0 or x=2/3
(More)

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Expert Answered

Asked 11/27/2012 1:04:48 PM

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12x^2 - 10x - 42 = 0

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Not Answered

Updated 48 days ago|7/30/2014 12:37:08 PM

1 Answer/Comment

12x^2 - 10x - 42 = 0

2(2x + 3)(3x - 7) = 0;

2(2x + 3) = 0; 4x + 6 = 0; 4x = -6; x = -6/4; x = -3/2

3x - 7 = 0; 3x = 7; x = 7/3

Solution: x = -3/2 or 7/3

2(2x + 3)(3x - 7) = 0;

2(2x + 3) = 0; 4x + 6 = 0; 4x = -6; x = -6/4; x = -3/2

3x - 7 = 0; 3x = 7; x = 7/3

Solution: x = -3/2 or 7/3

Added 48 days ago|7/30/2014 12:37:08 PM

x2 + 16x = 0 You can solve the above these three nethod by 1) factoring, 2) quadratic formula, and 3) completing the square

Weegy: You can solve it through 2) quadratic formula. Ask another one! (More)

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Expert Answered

Asked 11/28/2012 3:48:12 PM

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What do we do if the middle term of the quadratic equation (bx) is missing? Who wants to give the following a try?
2x2 + 1 = 0

Weegy: The syntax of quadratic equation that we are familiar with is ax^2+bx+c=0. This actually is an adfected quadratic equation.
The real form is ax^2+b=0, which is a pure quadratic equation.
2x^2+1=0
or, 2x^2=-1.
or, x^2=-1/2
or, x=sqrt(-1/2). (More)

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Asked 11/29/2012 12:10:50 AM

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