(287477464.18* x) + 486340506.11* x ^(0.5)= 3193310803.54 how do I solve for X?

Answer 1

Given:
287 477 464∙18(x) + 486 340 506∙11(x)^½ = 319 331 0803∙54
By converting the equation into a quadratic equation the quadratic formula may be used.
Let Z = X^½
Z² = X
287 477 464∙18(Z²) + 486 340 506∙11(Z) = 319 331 0803∙54
287 477 464∙18(Z²) + 486 340 506∙11(Z) - 319 331 0803∙54 = 0

Now, using the quadratic formula: -b±√(b²-4ac) / 2a

- 486 340 506∙11 ± √(486 340 506∙11 - (4)(287 477 464∙18)( - 319 331 0803∙54) / (2)(287 477 464∙18)

- 486 340 506∙11 ± √(3∙90854655x10^18) / (574 954 928∙2)


- 486 340 506∙11 ± 1977004465 / (574 954 928∙2)

Z = -246 334 4971 / 574 954 928∙2 or
Z = 149 066 3959 / 574 954 928∙2

Z = -4∙284414048 or Z = 2∙592662287

But Z² = X^½ = -4∙284414048 or 2∙592662287
→ Z² = X = 18∙25620373 or 6∙721867733
Now check these values with the equation.

287 477 464∙18(x) + 486 340 506∙11(x)^½ = 319 331 0803∙54
287 477 464∙18(18∙25620373) + 486 340 506∙11(-4∙284414048) = 319 331 0803∙54
52769944900 - 2083684096 = 319 331 0803∙54
Answer comfirmed.
X = 18∙25620373

Now check other factor:
287 477 464∙18(x) + 486 340 506∙11(x)^½ = 319 331 0803∙54
287 477 464∙18(6∙721867733) + 486 340 506∙11(2∙592662287) = 319 331 0803∙54
1932394114 [phone] = 3193310803.54
Answer comfirmed.
X = 6∙721867733

Answer 2

If your numbers are so large, what are you doing with those deimals? Shows u r not an engineer?

Say
ax + b x^.5 = c easiest way to do this would be to say that root x = u. then the equation becomes

au^2 + bu - c = 0 which is a quadratic equation with the roots

u = (-b + root(b^2 - 4 a (-c)) ) / 2a and -b - ....

u = (-b + root (b^2 + 4ac)) / 2a

substitute for a, b, c to get u

Then use x = u^2 to find x.

Hope this helps

Answer 3

1)square the whole equation
2)bring the (3193310803.54^2) to the left...as in subtract this value from both sides
3)u'll hav sth like
ax^2+bx-c=0
factorize this and u'll get sth like:
(x+p)(x+q)=0
4)therefore: (x+p)=0 or (x+q)=0
x=-p x=-q
NOTE**: the value of x CAN be positive ... because the value of p or q can also be nagative.

the numbers are long so it's gonna be pretty tedious... but this method is sure to work... and keep in mind tt since its a quadratic equation there will be 2 answers. (unless it's a perfest square and the 2 roots are common...or there are no roots...
but in this case:
ax^2+bx+c...
discriminant= b^2-4ac
=(486340506.11^2)^2-4(287477464.18^2)(3193310803.54^2)
... if this value is bigger than 0 than there are 2 final values of x
if this value is smaller than 0 than there are no final values of x
and if it is equal to 0 than there is only 1 answer

hope this helps =)

Answer 4

Square the equation. change sides so that right hand side is zero. Then solve as a quadratic equation e.g. ax^2+bx+c=0 -> x= {-b+/-(b^2-4ac)^(0.5)}/2a

Answer 5

Square the whole thing, then you're dealing with a simple quadratic. x = -b +- SQRT(b^2 - 4ac)/2a

And yeah, when you're dealing with numbers that large, nobody cares about the .18, they probably don't care about the last 5 digits. Unless of course you are at NASA, but I suppose you're not.... HOPE you're not.

Answer 6

Use a calculator; square your equation first.

Answer 7

use the 'abc' formula

x1,2 = (-b +/- sqrt(b^2 - 4ac))/2a

a = (287477464.18)^0.5
b = (486340506.11)^0.5
c = - 3193310803.54

x = (x1,2)^2 (modulo signs and errors)