find the exact solutions to 3x^2= 5x-1 by using the Quadratic Formula?

Answer 1

3x^2 -5x +1 =0 ..... eq (1)

Quadr. formula (Shri Daracharya's method): ax^2 +bx +c=0

gives x = - b ± sqrt (b^2 - 4.a.c) / 2.a

for eq (1)... x = 5 ± sqrt (5*5 - 4*3*1) / 2*3
= 5 ± sqrt (13) / 6
Thus, solutions are

6x1 = 5+sqrt (13), and
6x2 = 5- sqrt (13)

cross check:
here x1 + x2 (sum of roots) = 10/6=5/3 = -b/a,
and x1 * x2 (product) = 12/36=1/3 = c/a

Answer 2

a million. 4, -5/2 via factoring 2x2 - 3x - 20 = 0 to (2x+5)(x-4) = 0 and fixing each and every a million/2 (2x+5 = 0 and x-4 = 0). 2. it has 2 imaginary recommendations by way of fact the discriminant is detrimental so, "No authentic recommendations" is in all probability your answer 3. Use the quadratic formula to sparkling up for x while 8x2 + 7x - 8 = 0; you will get 2 imaginary recommendations. the respond is slightly messy to attempt to variety here. i think of it extremely is 7 +/- 3i squareroot of 23 throughout sixteen. 4. component to get x(x-12) = 0 and sparkling up the two areas; x = 0 and x - 12 = 0 so x = 12. the two 0 and 12 examine once you replace them back into the unique.

Answer 3

First putting the equation in proper form, you get 3x^2 - 5x + 1 = 0. The quadratic formua is x = (-b±√(b²-4ac))/2a. So a=3, b=-5, and c=1. Plugging these in to the formula, you get:

(-(-5) ± √((-5)²-4*3*1))) / (2*3)
(5 ± √(25 - 12)) / 6
(5 ± √(13)) / 6

So there are two solutions, x = (5+√13)/6 and x = (5-√13)/6

Answer 4

First: set the equation to "0" > subtract "5x" from both sides...

3x^2 - 5x = 5x - 5x - 1
3x^2 - 5x = - 1

*Add "1" to both sides...

3x^2 - 5x + 1 = - 1 + 1
3x^2 - 5x + 1 = 0

Sec: you need three coefficients to replace three variables...
a = 3 > b = - 5 and c = 1

Third: place the numbers in the quadratic formula which is...

x = [- b +/- V`(b^2 - 4ac)] / 2a

x = [- (-5) +/- V`((-5)^2 - 4(3)(1))] / 2(3)

x = [- 5 +/- V`((-5)^2 - 4(3)(1))] / 6

x = [- 5 +/- V`((5)(5) - 4(3)(1))] / 6

x = [- 5 +/- V`(25 - 4(3)(1))] / 6

x = [- 5 +/- V`(25 - 4(3))] / 6

x = [- 5 +/- V`(25 - 12)] / 6

x = [- 5 +/- V`(13)] / 6

*You have two solutions, one has addition-the other has subtraction.

a. x = [- 5 + V`(13)] / 6

x = -5/6 + (V`13)/6

b. x = [- 5 - V`(13)] / 6

x = - 5/6 - (V`13)/6

Answer 5

quadratic form is ax^2 + bx + c = 0
you have: 3x^2= 5x-1 so lets change it into the right form
3x^2= 5x-1
3x^2 - 5x + 1 = 0

so a = 3, b = -5, c = 1
just plug those into the quad formula and you got your answers :)

Answer 6

1st..put into standard form...so 3x^2-5x+1. rite? do u know the Q formula? it's -b (square root of) -b^2-4ac
---------------------------------
2a
so 3=a -5=b 1=c

5 (neg..cancels out) (square root of) 25-4(3)(1)
---------------------------------------------------------------
2(3)

5+-(plus or minus cuz it's a square root) 3.6
---------------------------------------------------------
6

so ur answers are= (3.6+5= 8.6...then divide by 6) = 1.43
= (3.6-5= -1.4..then divide by 6) = -.23

i'm pretty sure that's it.