Finding values from a equation?

Question

How do you find the values of x between -3 and 3 from the equation : y^2+xy+0.25=0

I need to draw a graph which shows the two solutions of the equation y^2 + xy+0.25 = 0
Can u please show and give the values of x between -3 and 3

Answer 1

Here's how to do it.

First check whether all real values of x are permitted. (They're not!)

Solving the quadratic equation for y, we find:

y = [-x +/- sqrt(x^2 -1)] / 2.

The sqrt factor shows that mod(x) can't be less than 1. This is essentially all you needed to know in order to demonstrate that only values of x satisfying 1 LE mod(x) LE 3 are allowed in the given range, where "LE" means "less than or equal to."

If that's all that you are interested in, we're finished. (The question didn't even ask for values of y associated with the allowed x's.)

Putting it in the simplest and most explicit terms --- in the range specified (-3 to +3), ALL values of x between -3 and -1, and between +1 and +3 (including those precise limits), are possible and permitted. Those between -1 and +1 (but not those precise limits) are NOT possible or permitted.

If, however, you have an enquiring mind, one can go further by showing that for x = +/-1, y = -/+1/2, and that dy/dx is "+/- infinity" at these values. Furthermore, for very large mod(x), the binomial expansion of the sqrt factor shows you that:

y ---> [- x +/- {x - 1/(2x)}] / 2 =

(i) - 1/(4x) + ... or (ii) - x + 1/(4x) + ...

These limits for large mod(x) plus the vertical tangents at x = +/-1, y = -/+ 1/2 enable one to quickly sketch the whole (x, y) relationship: two distorted hyperbola-like curves, reflections of one another through the origin. The one for positive x comes in from infinity just below the +x axis (value for large x ~ - 1/(4x)), passes vertically through the limiting x = 1 point found previously, and proceeds down and back to infinity, asymptotically tending to the line y = -x sloping down at - 45 degrees. (The other branch is simply the indicated reflection though the origin.)

Live long and prosper.

POSTSCRIPT So far, with 5 answers now in (including mine), no-one else has shown that the range mod(x) < 1 is EXCLUDED by the given relationship between x and y. The nearest anyone has come to this is the answer immediately below mine (by Andy). However, that only shows that whereas the (turning) points x = +/-1 are perfectly fine and real, x = 0 gives you a purely imaginary value for y. Such "sample evaluations" are not a really satisfactory way of establishing a "forbidden zone" for values of 'x'. My own argument, above, shows precisely what the forbidden zone is, and why.

SECOND POSTSCRIPT: PLEASE, SHIBZ, you now have to DO SOMETHING FOR YOURSELF! You have all the tools:

I've told you (and I'm the ONLY ONE WHO HAS TOLD YOU) that there's no value of x between -1 and +1 for which a real value of y exists. Others have shown you how to solve for y, given any OTHER value for x (without pointing out that some values ARE excluded). Some have evaluated real y-values for you, for ALLOWED BUT ONLY INTEGER values of x. I also verbally anticipated what you later asked in your "Additional details," by working out and describing the kind of (x, y) plot that you would obtain, for ALL allowed values of x. (You're only asked to plot the parts of such full curves for x between -3 and + 3.) I'll give you two more bits of information. For x = +3, the quadratic equation yields y = -3/2 +/- sqrt(2), so that the curve you will plot for positive x in the given range goes between (+3, -0.086... and (+3, -2.914...). [As stated earlier, the other branch is simply a reflection through the origin; simply make (x, y) --> (-x', -y').] You won't really need to go beyond the second significant decimal figure to draw a good looking curve.

WHAT MORE could you possibly and reasonably expect? There's a CONTINUOUS SET of x-values between (i) -3 and -1, and (ii) +1 and +3. CONTINUOUS means INFINITE ! Do you REALLY expect anyone to do an INFINITE AMOUNT OF WORK FOR YOU ??!!

Get some graph paper and a calculator, calculate some corresponding (x, y) values, and PLOT them.

Additional Hint: Once you know the y-range from the values of y for x = -3, -1, +1 and +3, it's easier (to get plotting values) to START WITH y in the relevant ranges, and CALCULATE x. THAT needs NO SOLUTION of a QUADRATIC EQUATION --- you simply re-write the defining equation as x = (explicit function of y). But do be a good guy/gal and now do some work yourself.

Answer 2

You replace the x with each value (-3, -2, -1, etc.) Then solve for y by the quadratic formula.

Example: when x is -3:

y^2 -3y + 0.25 = 0

y = 3 +- sqrt(9 - 4(1)(.25)) all over 2

=3 +- sqrt 8 all divided by 2

(do this on the calculator)

Answer 3

You can solve the equation assume x is a parameter.

For a quadratic equation like this:

ay^2 + by + c = 0

the solution is:

I will not bore you with the solution involving "completing the square" but you can see that in the attached reference.

y = [-b ± sqrt(b^2-4ac)]/2a

The given equation, a=1, b=x, c=0.25 (or 1/4)

So

y = [-x ± sqrt(x^2-1)]/2

So if you are just looking at integer values of x in the range [-3,3]:

x______x^2-1_________y
-3____8 = 2*4____(3/2)±sqrt(2) = (3/2)+sqrt(2) or (3/2)-sqrt(2)
-2______3______1±sqrt(3)/2 = 1+sqrt(3)/2 or 1-sqrt(3)/2
-1______0______1/2
0_____-1______±i/2 = +i/2 or -i/2
1______0______-1/2
2______3______-1±sqrt(3)/2 = -1+sqrt(3)/2 or -1-sqrt(3)/2
3______8______-(3/2)±sqrt(2) = -(3/2)+sqrt(2) or -(3/2)-sqrt(2)

I assume you know "i" is the imaginary number = sqrt(-1)

So you see using an approximation method and a calculator would not help much.

Answer 4

Hayharbr's answer makes no sense. Hayharbr showed how to find values for y, but you are asking for values of x.

x can be the sum of any two numbers whose product is .25. For example, the numbers 1 and .25 multiply to .25, so x can be 1.25. The resulting equation,

y^2 + 1.25y + .25 = 0

can be factored into

(y + 1)(y + .25) = 0

x can take on any value between -3 and 3, not just integer values as Hayharbr suggested. You don't find the values of x from the equation. You are told that x can be any number between -3 and 3, and that is the answer.

Answer 5

y^2+xy+0.25=0 rearange subtract y^2+.25 from each side
xy=-(y^2+.25) divide by y
x=-(y+.25/y) now substitute values for y & get x.
unless you neant find values for y when -3 in which case
y^2+xy+0.25=0 this is quadratic
t=(-x+/-√(x^2-1))/2 substitute values for x & solve.