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Gustavo Estes
Mathematics
20 May, 04:57
4x^ (2) = 13x-3 use quadratic formula
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David Garza
20 May, 05:23
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The quadratic formula is - b + / - sqrt (b^2-4ac) all over 2a.
First we have to get all the variables on one side so ... Subtract 4x: 0=-4x^ (2) + 13x-3
OR add 3 and subtract 13x: 4x^ (2) - 13x+3=0
Since I prefer a to be positive, I'm going to choose the second equation.
So ... now we just plug and chug. a is the value of the variable squared. In this case a=4. b is the value with the variable, or b=-13. c is the last term. c=3
**Remember: Ax^ (2) + By+C**
Now we just plug everything in.
-b = 13 (negative minus a negative is a positive)
+/-sqrt ((-13) ^ (2) - 4 (4) (3))
all over 2 (4)
So work with the radical first.
(-13) ^2=169
4 (4) (3) = 48
+/-sqrt (169-48)
+-sqrt (121)
sqrt (121) = 11
Now it's just: (13+/-11) / 2 (4)
(13+/-11) / 8
Split this into two equations:
(13+11) / 8
(13-11) / 8
Solve both: 24/8=3
2/8=1/4
So x = 3, 1/4
Plug them back in and see if there's one solution or two:
4 (3) ^2=13 (3) - 3
36=36
So x=3.
How about 1/4?:
4 (1/4) ^2=13 (1/4) - 3
4 (1/16) = 13/4-3
4/16=13/4-3
1/4=13/4-3
1/4=13/4 - (3x4) / (1x4) * Like denominators to add or subtract*
1/4=13/4-12/4
1/4=1/4.
So x=1/4.
In this case, both answers work. So the answer, using the quadratic formula is x=1/4, x=3
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