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how do you solve 3x^2 - 4x - 5= 0 using the quadratic formula? and leaving irrational roots in simplest form?????

2006-08-31 06:35:02 · 6 answers · asked by latrice7437 2 in Science & Mathematics Mathematics

6 answers

well dear f(x) = 3x^2 - 4x - 5
in general form we have
ax^2 + +bx + c =0

so if a = +3 , b= -4 & c = -5

∆ = b^2 - 4ac
∆ = (-4^2) - 4* 3* -5 = 16 +60= 76

now
x1 = (-b + √∆) / 2a = ( -(-4) + √76 )/ 2*3 = 4+7.45 / 6
x1 = (-b - √∆) / 2a = ( -(-4) - √76 )/ 2*3 = 4- 7.45 / 6

2006-08-31 11:05:15 · answer #1 · answered by sweetie 5 · 1 0

To use the quadratic formula, your equation must be in the form:
ax² + bx + c = 0

Your equation is already in this form
3x² + (-4)x + (-5) = 0

So you can easily determine a, b and c (3, -4, -5).

Now you just plug that into the quadratic formula. It's hard to type this all on a single line here and make it readable, but it looks something like this:

x = [ -b ± √(b²-4ac) ] / 2a

If you were writing this on your paper, you would make a fraction with -b ± √(b²-4ac) in the numerator and 2a in the denominator.

Now, just plug in the values above for a = 3, b = -4, c = -5

Numerator: -b ± √(b²-4ac) = -(-4) ± √( (-4)² - 4(3)(-5) )
Denominator: 2a = 2(3)

Numerator: 4 ± √(16 + 60)
Denominator: 6

Numerator 4 ± √76
Denominator: 6

Now √76 can be simplified by factoring out a 4:
Numerator 4 ± √4 * √19
Denominator: 6

Numerator 4 ± 2 * √19
Denominator: 6

Divide top and bottom by 2:
Numerator 2 ± √19
Denominator: 3


So the answers are:
x = (2 + √19) / 3
x = (2 - √19) / 3

Or simply:
x = (2 ± √19) / 3

The decimal values are:
x ≈ 2.120... or -0.7863...

2006-08-31 13:49:27 · answer #2 · answered by Puzzling 7 · 0 1

3x²-4x-5=0 is in the standard form ax²+bx+c=0 so it's
plug 'n play
x = (-b±√(b²-4ac))/2a and
x = (4±√(16+60))/6) = (4±√76)/8 = 1/2 ± √19/4


Doug

2006-08-31 13:48:32 · answer #3 · answered by doug_donaghue 7 · 0 0

You have to break the equation up into a, b & c.

In your case:

a = 3
b = -4
c = -5

Then plug them into the quadratic formula found here (it may be confusing to type it out):
http://mathworld.wolfram.com/QuadraticFormula.html

Use the one that is all over "2a"

Simply plug in the values and solve it.

2006-08-31 13:44:26 · answer #4 · answered by dreft 2 · 0 0

a=3 b=(-4) c=(-5)

(-b) +/- (sq root(b^2-(4ac))) divided by (2a)

(-(-4)) +/- (sq root(76)) divided by 6
4 +/- 2(sq root(19))/6
(2 +/- (sq root(19)))/3

about 2.12 or (-.79)

2006-08-31 13:44:40 · answer #5 · answered by Anonymous · 0 0

Do your own homework!

2006-08-31 13:47:00 · answer #6 · answered by Anonymous · 0 1

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