Help me solve some algebra equations

[isuranja]

y^2-8xy-x^2=0

I got model question in a paper that,

1) Prove the identity (x² + y²) - 64x²y² ≡ (x^4 + y^4) - 62x²y²
Hence find the factors (x^4 + y^4) - 62x²y² = -2135
x²+8xy+y = 61

2) solve the equation √x-3 + √2x+1 = 2√x

I am confusing of this please help on this.

 

[mayyan]

Re: problem in algebra

1) x^2 + y^2 - 64(x^2)*(y^2) = x^4 + y^4 - 62(x^2)*(y^2)
x^2 + y^2 = x^4 + y^4 + 2(x^2)*(y^2)
x^2 + y^2 = (x^2 + y^2)^2

now let a:= x^2 + y^2 then:

a = a^2
a^2 - a = 0
a(a-1) = 0 => a=0 or a=1
=> x^2 + y^2=0 or x^2 + y^2=1
which means the equation doesnt hold for any x,y.

2) √(x-3) + √(2x+1) = 2√x

trems of existens:
x≥3 and x≥-0.5 and x≥0 => x≥3

√(x-3) + √(2x+1) = 2√x / ( )²
x-3 + 2√(x-3)√(2x+1)+ 2x+1 = 4x
2√(x-3)√(2x+1) = x+2 / ( )²
8x² - 20x - 12 = x² +2x + 4
7x² - 22x - 16 = 0

....



hope i got it right

 

[steve10]

Re: problem in algebra

I am amazed by your understanding capability. Could you please explain what he means when he asks to prove the "identity" (x² + y²) - 64x²y² ≡ (x^4 + y^4) - 62x²y² ? My understanding about "identity" is that the equation holds for whatever (x,y) you choose, but it is not the case here. So .... what do you mean when you think that is an identity?

 

[mayyan]

Re: problem in algebra

I am amazed by your understanding capability. Could you please explain what he means when he asks to prove the "identity" (x² + y²) - 64x²y² ≡ (x^4 + y^4) - 62x²y² ? My understanding about "identity" is that the equation holds for whatever (x,y) you choose, but it is not the case here. So .... what do you mean when you think that is an identity?

My understanding of identity is same as yours. I didnt really knew what he wanted to do in the first question in particular i didnt understand his explanation. I figured that the original question could be whether this is an identity or to find where the equation holds. I proved the firt, thinking that if its the second its to hard to continue from where i stoped.

Anyway i didnt get any response from him so i dont know if i got it right.

 

[tkbits]

Re: problem in algebra

(x² + y²) - 64x²y² ≡ (x^4 + y^4) - 62x²y² is not an identity, but
(x² + y²)² - 64x²y² ≡ (x^4 + y^4) - 62x²y² is an identity.

 

[mokong]

Re: problem in algebra

(a)
tkbits is right: (x^2 + y^2)^2 = x^4 + 2 x^2 y^2 + y^4, so adding -64 x^2 y^2 to both sides gives (x^2 + y^2)^2 - 64 x^2 y^2 = x^4 + y^4 - 62 x^2 y^2.

I don't really understand "Hence find the factors (x^4 + y^4) - 62x²y² = -2135
x²+8xy+y = 61 " but I assume those are given conditions and we are to find x and y.

From above, (x^4 + y^4) - 62x²y² = -2135 means (x^2 + y^2)^2 - 64 x^2 y^2 = (x^2 + y^2)^2 - (8 x y)^2 = (x^2 + y^2 - 8xy)(x^2 + y^2 + 8xy) = -2135. But x^2 + 8xy + y^2 = 61 so we have (x^2 + y^2 - 8xy) * 61 = -2135 or x^2 + y^2 - 8xy = -2135/61 = -35. So far we have

x^2 + 8xy + y^2 = 61
x^2 - 8xy + y^2 = -35.

This is 2 equations in 2 unknowns: it is solvable. Add the first to the second to get 2*(x^2 + y^2) = 61 - 35 = 26 or x^2 + y^2 = 13. Subtract the second from the first to get 8xy - (-8xy) = 61 - (-35) or 16xy = 96 or finally xy = 6. Now we have

x^2 + y^2 = 13
xy = 6

It is easy to guess from the first equation that x=2, y=3 and x=-2, y=-3 are solutions (as are x=3,y=2 and x=-3,y=-2). The second equation verifies this. If you want to be more rigorous about it, use the fact that y = 6/x to rewrite the first equation:

x^2 + 36/x^2 = 13
so x^4 - 13x^2 + 36 = 0
or (x^2 - 9)(x^2 - 4) = 0

Thus either x^2 - 9 = 0, giving x = +/- 3, or x^2 - 4 = 0, giving x = +/- 2. After solving the corresponding y (namely, y = 6/x) these are (of course) similar to our guesses above.

(b)
mayyans solution is correct up to this part: 2√(x-3)√(2x+1) = x+2. Square both sides to get 4(x-3)(2x+1) = (x+2)^2, so 8x^2 - 20x - 12 = x^2 + 4x + 4. After transposing we have the quadratic eqn:

7x² - 24x - 16 = 0

We can factor the left to get

(7x + 4)(x - 4) = 0.

Here, as mayyan pointed out, we have a constraint for x: x≥3. This is in order for our original equation to make sense in the REAL numbers. We impose this on the roots x = -4/7 and x = 4 of the preceeding quadratic equation. It turns out that only x = 4 (which is >= 3) IS acceptable.

 

[hamidr_karami]

Re: problem in algebra

hi

solve (x-3)^0.5+(2*x+1)^0.5-2*(x)^0.5=0
x= 4

bye

 

[hanglb]

problem in algebra

The problem involves the solutions of nolinear algebra equations. You can use groebner basis to solve it.