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Questions asked by the same visitor
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 21 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
Added 21 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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