Precalculus need help!!
posted by Isis
Quadratic Formula
1.) x+6/x=7
2.) 4x/x+4 + 3/x1 = 15/x^2+3x4
3.) x+3/x  2/x+3 = 6/x^2+3x
formula:
B+ sqrt B^2 + (4)(a)(c)/2a
f u only wanted to use the formula it's not necessary though.
please help me! it's going to pass tomorrow!! please!

Reiny
your formula should say:
x = (b ± √(b^2  4ac) )/(2a)
1. x + 6/x = 7
each term by x
x^2 + 6 = 7x
x^2 + 7x + 6 = 0
it factors, no need for the formula
(x+1)(x+6) = 0
x = 1 or x = 6
2. you probably meant:
4x/(x+4) + 3/(x1) = 15/(x^2 + 3x  4)
4x/(x+4) + 3/(x1) = 15/((x1)(x+4) )
multiply each term by (x+4)(x1) , the LCD
4x(x1) + 3(x+4) = 15
4x^2  4 + 3x + 12 = 15
4x^2 + 3x 7 = 0
x = (3 ±√(94(4)(7)) )/8
= (3 ±√121)/8 = (3 ± 11)/8
x = 1 or x = 7/4
or.... we could have factored it again
(x1)(4x + 7) = 0
x = 1 or x = 7/4
3. again, brackets will be essential.
(x+3)/x  2/(x+3) = 6/( x(x+3))
the LCD is x(x+3) , so multiply each term by x(x+3)
You can take it from there.
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