Equations...?

Question

If x1 is radical of the equation:
5x^3 - 6 = 0
and x2 the radical of the equation:
6x^3 -5 = 0
prove that: x1 + x2 > 2

Answer 1

5x^3 - 6 = 0, x1 is the solution

6x^3 - 5 = 0, x2 is the solution

Adding them together,

We can actually calculate the real solutions directly.

5x^3 - 6 = 0 implies
5x^3 = 6, which further implies
x^3 = 6/5, so
x = (6/5)^(1/3)
Therefore, x1 = (6/5)^(1/3)

6x^3 - 5 = 0
6x^3 = 5
x^3 = (5/6)
x = (5/6)^(1/3)
Therefore, x2 = (5/6)^(1/3)

x1 + x2 = (6/5)^(1/3) + (5/6)^(1/3)

For (6/5)^(1/3) + (5/6)^(1/3), what I'm going to do is rationalize this by multiplying the necessary value to obtain (6/5) + (5/6).
Remember that a sum of cubes a^3 + b^3 factors as
(a + b)(a^2 - ab + b^2), so what we're going to do is multiply top and bottom by [(6/5)^(1/3)]^2 - [(6/5)^(1/3)][(5/6)^(1/3)] + [(5/6)^(1/3)]^2. By "top and bottom", we're going to do this with the imaginary denominator that is equal to 1.

x1 + x2 = [ (6/5) + (5/6) ] / [ (6/5)^(2/3) - [(6/5)^(1/3)][(5/6)^(1/3)] + (5/6)^(2/3)

We can rewrite (6/5)^(1/3) * (5/6)^(1/3) as (6/5 * 5/6)^(1/3), which is equal to 1^(1/3), which is equal to 1, giving us

x1 + x2 =[ (6/5) + (5/6) ] / [ (6/5)^(2/3) - 1 + (5/6)^(2/3) ]

Simplifying the top,

x1 + x2 = [ (36/30) + (25/30) ] / [ (6/5)^(2/3) - 1 + (5/6)^(2/3) ]
x1 + x2 = [ (61/30) ] / [ (6/5)^(2/3) - 1 + (5/6)^(2/3) ]

Got stuck at this point... I might still work on this problem later.

One thing to note is that

(6/5)^(1/3) > (5/6)^(1/3)

Moving the (5/6)^(1/3) over to the left hand side, we get

(6/5)^(1/3) - (5/6)^(1/3) > 0

Squaring both sides,

[(6/5)^(1/3)]^2 - 2 (6/5)^(1/3) (5/6)^(1/3) + [(5/6)^(1/3)]^2 > 0

[(6/5)^(1/3)]^2 - 2 + [(5/6)^(1/3)]^2 > 0
[(6/5)^(1/3)]^2 + [(5/6)^(1/3)]^2 > 2, OR

[(6/5)^(2/3)] + [(5/6)^(2/3)] > 2

Remember that

x1 + x2 = [ (6/5) + (5/6) ] / [ (6/5)^(2/3) - 1 + (5/6)^(2/3) ]

Rearranging the denominator,

x1 + x2 = [ (6/5) + (5/6) ] / [ (6/5)^(2/3) + (5/6)^(2/3) - 1 ]

But, look at the denominator; we have [(6/5)^(2/3) + (5/6)^(2/3)], and we *know* this is greater than 2. Therefore,

[ (6/5) + (5/6) ] / [ (6/5)^(2/3) + (5/6)^(2/3) - 1 ] >
[ (6/5) + (5/6) ] / (2 - 1)

[ (6/5) + (5/6) ] / [ (6/5)^(2/3) + (5/6)^(2/3) - 1 ] > [ (6/5) + (5/6) ]

Therefore, by transitivity,

x1 + x2 > [(6/5) + (5/6)]

Simplifying the right hand side,

x1 + x2 > [36/30 + 25/30]
x1 + x2 > [61/30]

Note that 61/30 is greater than 2 (since 60/30 is equal to 2). Therefore,

61/30 > 2. By transitivity,

x1 + x2 > 2

Answer 2

because you've x AND x² interior the equation someplace, this is a quadratic. you pick to get it interior the variety: ax² + bx + c = 0 So our equation is: x² - 7x + 38 = 5x + 3 We really only pick each and every thing on the left side of the equation, so enable's subtract 5x on both side: x² - 12x + 38 = 3 Now we subtract 3 on both side: x² - 12x + 35 = 0 we've a quadratic expression in which a = a million, b = -12, and c = 35. There are 2 trouble-free strategies of fixing. a million) Factoring: (x - 7)(x - 5) = 0 x = 7 or 5 2) Quadratic formulation x = (-b ±?(b² - 4ac)) / 2a = (12±?((-12)² + (-4)(a million)(35)) / (2*a million) =(12±?(one hundred and forty four - one hundred and forty))/2 =(12±?(4))/2 = (12+2)/2 or (12-2)/2 = 14/2 or 10/2 = 7 or 5

Answer 3

well kitty lets make it simpler

say x1 = a and x2 = b

just solve for a and b


5a^3 - 6 = 0

solve for a

a = cube root of 6/5 = 1.0626585691826111

then we do the same for b

6b^3 - 5 = 0

solve for b

b = cube root of 5/6 = 0.941036028880903


when you add a and b you get 2.0036945980636397

just slightly above 2

hope that helps

Answer 4

A=B
B=A
X=N
N*X^2(1002/100)=12b*c/(1500M*x+y)^2
(XZYBSNB/2^PQLMNA^XX%$VM)=0
THAT MEANS WORLD EQUAL ZERO

Answer 5

try studing and know i will not help u on ur quiz