Could you just help me solving this maths problem?

Question

6x^2 + (sqrt(3) - 13)x +3 + sqrt(3) = 0

Well, folks, the point in that problem is that I should keep the squire roots and use the properties of the roots.
When I estimate the D, I find, that it is 100- 50*sqrt(3), but I cannot put that under a sq root.

Answer 1

.

This is jus another quadratic equation... The co-efficients are jus irrational... Nevertheless, this equation gives 2 perfectly real roots...

The 2 roots are:
1.7090853153992358558683710732327
0.168906216672617928543721203182

.

Answer 2

No problem. So you end up with :
x = {13 - sqrt(3) ± sqrt[100 - 50*sqrt(3)]} / 12

Let's look at sqrt[100 - 50*sqrt(3)].
Taking out the factor 25, this becomes :
sqrt{25[4 - 2*sqrt(3)]}, which is 5*sqrt[4 - 2*sqrt(3)].

Now let's look at sqrt[4 - 2*sqrt(3)].
Assume that sqrt[4 - 2*sqrt(3)] = sqrt(a) - sqrt(b).

Squaring both sides gives :
4 - 2*sqrt(3) = a + b - 2*sqrt(ab)

Equating rationals and irrationals gives :

4 = a + b
3 = ab

So, b = 4 - a.
Plug this into 3 = ab to get 3 = a(4 - a)
Rearranging gives : a^2 - 4a + 3 = 0
Factorising gives : (a - 3)(a - 1) = 0
Thus, a = 3 or 1, so, b = 1 or 3.
It doesn't matter which is which,
but we'll take a = 3 and b = 1,
so that we get positive values for the equation.

Therefore, sqrt[4 - 2*sqrt(3)] = sqrt(3) - 1

Now, x will be {13 - sqrt(3) ± 5[sqrt(3) - 1)] / 12
Now we can add them.

Thus, x = [2 + sqrt(3)] / 3 or x = [3 - sqrt(3)] / 2.

I haven't tested these by plugging them back
into the equation, so I'll leave that for you.

Answer 3

The QuadraticFormula is used to solve equations like:

a x^2 + b x + c = 0 .


It gives the possible values for x in terms of a, b, and c. You can derive it by completing the square; the result is this:

If a x^2 + b x + c = 0 then either


- b + sqrt( b^2 - 4 a c )
x = --------------------------- or
2 a


- b - sqrt( b^2 - 4 a c )
x = ---------------------------
2 a

Answer 4

6x^2 + (sqrt(3) - 13)x +3 + sqrt(3) = 0
6x^2 + sqrt(3)x - 13x +3 + sqrt(3) = 0
6x^2 + sqrt(3)x - 13x +3 + sqrt(3) = 0
x(6x + sqrt(3) -13) +3 + sqrt(3) = 0
hell i give up...too tired

Answer 5

use discrete methode
i.e.
d=b^2-4ac and d=0 for real roots
a,b,c are
ax^2+bx+c=0
for any quadratic equation

Answer 6

it is a quadratic equation, u can use completing square method or use the quadratic formula, you can get two roots i think!

Answer 7

How much wood would a woodchuck chuck if a woodchuck could chuck wood?

Answer 8

You need to do your own homework honey -- that's how you learn!

Answer 9

x=x

Answer 10

.63397 or 1.244