Quadratic equations solved by factoring & with tables.
I missed a day of school & my teacher wouldn't explain to me how to solve it
I missed a day of school & my teacher wouldn't explain to me how to solve it
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2x^2 - x - 3 = 0
(2x + 1)(x - 3) = 0
x - 1/2 or x = 3
(2x + 1)(x - 3) = 0
x - 1/2 or x = 3
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formula for a quadratic: ax²+bx+c=0
2x²-x=3 --> if you subtract 3 from both sides the resulting equation is 2x²-x-3=0. This can then be factored into (2x-3)(x+1)=0 because -3 and 1 are factors of -3 (c) and when you do -3*1 + 2*1 (the outer and inner parts of the factored equation) you get -1 (b). You can then assume that 2x-3=0 and x+1=0 because 0*something = 0. I'm assuming you can do the rest from there
2x²-x=3 --> if you subtract 3 from both sides the resulting equation is 2x²-x-3=0. This can then be factored into (2x-3)(x+1)=0 because -3 and 1 are factors of -3 (c) and when you do -3*1 + 2*1 (the outer and inner parts of the factored equation) you get -1 (b). You can then assume that 2x-3=0 and x+1=0 because 0*something = 0. I'm assuming you can do the rest from there
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Dear Angel,
To convert to a quadratic form, subtract -3 from both side, to make it equal to zero.
2x^(2)-x-3=0
For a polynomial of the form ax^(2)+bx+c, find two factors of a*c (-6) that add up to b (-1).In this problem 1*-(3)/(2)=-(3)/(2) (which is (c)/(a)) and 1-(3)/(2)=-(1)/(2) (which is ((b)/(a)) , so insert 1 as the right hand term of one factor and -(3)/(2) as the right-hand term of the other factor.
(x+1)(x-(3)/(2))=0
Remove the fraction by multiplying the first term of the factor by the denominator of the second term.
(x+1)(2x-3)=0
Set each of the factors of the left-hand side of the equation equal to 0.
x+1=0_2x-3=0
Since 1 does not contain the variable to solve for, move it to the right-hand side of the equation by subtracting 1 from both sides.
x=-1_2x-3=0
Set each of the factors of the left-hand side of the equation equal to 0.
x=-1_2x-3=0
Since -3 does not contain the variable to solve for, move it to the right-hand side of the equation by adding 3 to both sides.
x=-1_2x=3
Divide each term in the equation by 2.
x=-1_(2x)/(2)=(3)/(2)
Simplify the left-hand side of the equation by canceling the common factors.
x=-1_x=(3)/(2)
The complete solution is the set of the individual solutions.
x=-1,(3)/(2)
===================
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To convert to a quadratic form, subtract -3 from both side, to make it equal to zero.
2x^(2)-x-3=0
For a polynomial of the form ax^(2)+bx+c, find two factors of a*c (-6) that add up to b (-1).In this problem 1*-(3)/(2)=-(3)/(2) (which is (c)/(a)) and 1-(3)/(2)=-(1)/(2) (which is ((b)/(a)) , so insert 1 as the right hand term of one factor and -(3)/(2) as the right-hand term of the other factor.
(x+1)(x-(3)/(2))=0
Remove the fraction by multiplying the first term of the factor by the denominator of the second term.
(x+1)(2x-3)=0
Set each of the factors of the left-hand side of the equation equal to 0.
x+1=0_2x-3=0
Since 1 does not contain the variable to solve for, move it to the right-hand side of the equation by subtracting 1 from both sides.
x=-1_2x-3=0
Set each of the factors of the left-hand side of the equation equal to 0.
x=-1_2x-3=0
Since -3 does not contain the variable to solve for, move it to the right-hand side of the equation by adding 3 to both sides.
x=-1_2x=3
Divide each term in the equation by 2.
x=-1_(2x)/(2)=(3)/(2)
Simplify the left-hand side of the equation by canceling the common factors.
x=-1_x=(3)/(2)
The complete solution is the set of the individual solutions.
x=-1,(3)/(2)
===================
Who am I, see all my videos on Youtube...
http://www.youtube.com/results?search_qu…
THANKS
===================
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4
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2x² - x = 3
2x² - x - 3 = 0
x = [1 ± √(1² - 4·2·(-3))]/(2·2)
= [1 ± √25)]/4
= [1 ± 5]/4
= -1, 1.5
2x² - x - 3 = 0
x = [1 ± √(1² - 4·2·(-3))]/(2·2)
= [1 ± √25)]/4
= [1 ± 5]/4
= -1, 1.5
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2x²-x=3
2x²-x-3=0
(2x-3)(x+1)=0
2x-3=0
2x=3
x=3/2
x+1=0
x=-1
2x²-x-3=0
(2x-3)(x+1)=0
2x-3=0
2x=3
x=3/2
x+1=0
x=-1