For some positive real numbers x and y , x 4 − 4 x 2 y + 3 x 2 y 2 + 5 x 2 − 1 2 y 3 + 1 5 y 2 = 0 . What is the value of 4 y − x 2 ?
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hi MSTang!
where did the exponent 4 go?
The given equation can be factorized as follows : x 4 − 4 x 2 y + 3 x 2 y 2 + 5 x 2 − 1 2 y 3 + 1 5 y 2
= ( x 4 − 4 x 2 y ) + ( 3 x 2 y 2 − 1 2 y 3 ) + ( 5 x 2 + 1 5 y 2 )
= x 2 ( x 2 − 4 y ) + 3 y 2 ( x 2 − 4 y ) + 5 ( x 2 + 3 y 2 )
= ( x 2 + 3 y 2 ) ( x 2 − 4 y ) + 5 ( x 2 + 3 y 2 )
= ( x 2 + 3 y 2 ) ( x 2 − 4 y + 5 )
Since the given equation equals zero , we have
( x 2 + 3 y 2 ) ( x 2 − 4 y + 5 ) = 0 which means either x 2 + 3 y 2 = 0 or x 2 − 4 y + 5 = 0 . Considering the second condition , we get x 2 − 4 y + 5 = 0 ⇒ 4 y − x 2 = 5
0 = ( x 4 − 4 x 2 y + 5 x 2 ) + ( 3 x 2 y 2 − 1 2 y 3 + 1 5 y 2 )
0 = x 2 ( x 2 − 4 y + 5 ) + 3 y 2 ( x 2 − 4 y + 5 )
0 = ( x 2 + 3 y 2 ) ( x 2 − 4 y + 5 )
x 2 − 4 y + 5 = 0
4 y − x 2 = 5
You need to explain why we must have x 2 − 4 y + 5 = 0 . Why can't x 2 + 3 y 2 = 0 ?
thanks..
Note that
x 4 − 4 x 2 y + 3 x 2 y 2 + 5 x 2 − 1 2 y 3 + 1 5 y 2 ( x 2 − 4 y + 5 ) ( x 2 + 3 y 2 ) = = 0 .
We know that x and y are strictly positive, so
x 2 + 3 y 2 > 0 ⟹ x 2 − 4 y + 5 = 0 ,
hence
4 y − x 2 = 5 .
x 4 − 4 x 2 y + 3 x 2 y 2 + 5 x 2 − 1 2 y 3 + 1 5 y 2 = 0
x 4 + 3 x 2 y 2 + 5 x 2 − 4 x 2 y + 1 5 y 2 − 1 2 y 3 = 0
x 2 ( x 2 + 3 y 2 ) + x 2 ( 5 − 4 y ) + 3 y 2 ( 5 − 4 y ) = 0
x 2 ( x 2 + 3 y 2 ) + ( 5 − 4 y ) ( x 2 + 3 y 2 ) = 0
( x 2 + 3 y 2 ) ( x 2 − 4 y + 5 ) = 0
x 2 − 4 y + 5 = 0
4 y − x 2 = 5
The goal will be to factorize the expression into something with 4 y − x 2 as a factor.
Therefore, x 4 − 4 x 2 y + 3 x 2 y 2 + 5 x 2 − 1 2 y 3 + 1 5 y 2 = − x 2 ( 4 y − x 2 ) − 3 y 2 ( 4 y − x 2 ) + 5 ( x 2 + 3 y 2 ) = − ( 4 y − x 2 ) ( x 2 + 3 y 2 ) + 5 ( x 2 + 3 y 2 ) = ( x 2 − 4 y + 5 ) ( x 2 + 3 y 2 ) = 0
Since x and y are positive real numbers, both x 2 and 3 y 2 will be positive. Thus, for the entire expression to be equals to 0 , x 2 − 4 y + 5 = 0 . Therefore, x 2 − 4 y = − 5 , and 4 y − x 2 = 5
x 4 − 4 x 2 y + 3 x 2 y 2 + 5 x 2 − 1 2 y 3 + 1 5 y 2 = 0
can reduce to
( x 2 + 3 y 2 ) ( 4 y − x 2 − 5 ) = 0
Therefore
So 4 y − x 2 = 5 .
Always remember to explain why you have rejected possible solutions. Otherwise the validity of your steps will be brought into question.
For this problem, it is important to explain why x 2 + 3 y 2 = 0 .
Rearranging the equation, you can have x 4 + 3 x 2 y 2 − 4 x 2 y − 1 2 y 3 + 5 x 2 + 1 5 y 2 = 0 . Then, by factoring you can have x 2 ( x 2 + 3 y 2 ) − 4 y ( x 2 + 3 y 2 ) + 5 ( x 2 + 3 y 2 ) = 0 . Hence, ( x 2 − 4 y + 5 ) ( x 2 + 3 y 2 ) = 0 . x 2 − 4 y + 5 = 0 . x 2 − 4 y + = − 5 . 4 y − x 2 = 5
Factoring the given equation will result to:
( x 2 + 3 y 2 ) ( x 2 − 4 y + 5 ) = 0 .
Note that since x and y are both positive real numbers, the first factor does not make the product zero.
Therefore x 2 − 4 y + 5 = 0 . Rearranging the equation will give us 4 y − x 2 = 5
Therefore the answer is 5 .
The equation is equivalent to ( x 2 + 3 y 2 ) ( 5 − 4 y + x 2 ) = 0 . Since $x,y>0$ then 5 − 4 y + x 2 = 0 or x 2 − 4 y = 5 .
Let us modify the expression a bit:
x 4 − 4 x 2 y + 3 x 2 y 2 + 5 x 2 − 1 2 y 3 + 1 5 y 2 = 0 ⇒
⇒ − x 2 ( 4 y − x 2 ) − 3 y 2 ( 4 y − x 2 ) + 5 ( x 2 + 3 y 2 ) = 0 ⇒
⇒ − ( x 2 + 3 y 2 ) ( 4 y − x 2 ) + 5 ( x 2 + 3 y 2 ) = 0 ⇒
⇒ − ( x 2 + 3 y 2 ) ( 4 y − x 2 ) = − 5 ( x 2 + 3 y 2 ) ⇒
⇒ 4 y − x 2 = − ( x 2 + 3 y 2 ) − 5 ( x 2 + 3 y 2 ) ⇒
⇒ 4 y − x 2 = 5
From this, we can clearly see that 4 y − x 2 equals exactly 5 , which is the correct answer.
Here we need to find the value of 4 y − x 2 , So let 4 y − x 2 = k ⇒ x 2 = 4 y − k . Putting this in the given equation we get, ( 4 y − k ) 2 − 4 ( 4 y − k ) y + 3 ( 4 y − k ) 2 y 2 + 5 ( 4 y − k ) − 1 2 y 3 + 1 5 y 2 = 0 which on simplifying reduces to k 2 − 4 y k − 3 y 2 k + 2 0 y − 5 k + 1 5 y 2 = 0 ⇒ 3 ( 5 − k ) y 2 + 4 ( 5 − k ) y + k ( k − 5 ) = 0 ⇒ ( 5 − k ) ( 3 y 2 + 4 y − k ) = 0 . Therefore we get k = 5 . Hence 4y-x^2=5
x^2(x^2 - 4y) + 3y^2(x^2 - 4y) + 5(x^2 + 3y^2) = 0 (x^2 + 3y^2)(x^2 - 4y) + 5(x^2 + 3y^2) = 0 (x^2 - 4y + 5)(x^2 + 3y^2) = 0 x^2 - 4y + 5 = 0 4y - x^2 = 5
This problem can be solved with factorization:
x 4 − 4 x 2 y + 3 x 2 y 2 + 5 x 2 − 1 2 y 3 + 1 5 y 2 = 0 3 x 2 y 2 + 5 x 2 − 1 2 y 3 + 1 5 y 2 = x 2 ( 4 y − x 2 ) 3 y 2 ( x 2 − 4 y ) + 5 ( x 2 + 3 y 2 ) = x 2 ( 4 y − x 2 ) 3 y 2 + ( x 2 − 4 y ) 5 ( x 2 + 3 y 2 ) = − x 2 ( x 2 − 4 y ) 5 ( x 2 + 3 y 2 ) = − ( x 2 + 3 y 2 ) 5 = − ( x 2 − 4 y ) 5 = 4 y − x 2
x^4-4x^2y+3x^2y^2+5x^2-12y^3+15y^2=0,
x^4-4x^2y+5x^2 + 3x^2y^2-12y^3+15y^2=0,
x^2(x^2-4y+5)+3y^2(x^2-4y+5)=0,
(x^2+3y^2)(x^2-4y+5)=0,
x^2-4y+5=0,
or
4y-x^2=5
This expression can be factorised as (x^2+3y^2)(x^2-4y+5)=0 Therefore making x^2-4y+5=0 We get 4y-x^2=5
Notice that we can factor the left-hand side of the equation as follows: x4−4x2y+3x2y2+5x2−12y3+15y2=x2(x2−4y)+3x2y2−12y3+5x2+15y2=x2(x2−4y)+3y2(x2−4y)+5x2+15y2=(x2+3y2)(x2−4y)+5(x2+3y2)=(x2+3y2)(x2−4y+5). Therefore, either x2+3y2=0 or x2−4y+5=0. However, since x and y are positive, x2+3y2>0, so it must be the case that x2−4y+5=0. Therefore, 4y−x2=5.
We rewrite the equation as ( x 2 − 4 y + 5 ) ( x 2 + 3 y 2 ) = 0 and therefore x 2 − 4 y + 5 = 0 or 4 y − x 2 = 5
First, simplify or factor the equation..
( x 4 + 3 x 2 y 2 ) + ( 5 x 2 + 15 y 2 ) - ( 4 x 2 y - 12 y 3 ) = 0
then remove the common factor;
x 2 ( x 2 + 3 y 2 ) + 5( x 2 + 3 y 2 ) - 4y( x 2 + 3 y 2 = 0
and since the 3 has common factor, we can unite them using distributive property;
( x 2 - 4y + 5)( x 2 + 3 y 2 ) = 0
and the two factors can be equate to zero (but not both equal to zero)
we're looking for the value of ( 4y - x 2 );
( \\\( x^\{2\} \\\) - 4y + 5)
=> 4y - \\\( x^\{2\} \\\) = 5
By the Commutative Property, this equation also means x 4 − 4 x 2 y + 3 x 2 y 2 − 1 2 y 3 + 5 x 2 + 1 5 y 2 = 0 . Simplifying, this equals x 2 ( x 2 − 4 y ) + 3 y 2 ( x 2 − 4 y ) + 5 ( x 2 + 3 y 2 ) = ( x 2 + 3 y 2 ) ( x 2 − 4 y ) + 5 ( x 2 + 3 y 2 ) = 0 . Therefore, ( x 2 + 3 y 2 ) ( x 2 − 4 y ) = − 5 ( x 2 + 3 y 2 ) , and x 2 − 4 y = − 5 . Finally, 4 y − x 2 = 5 .
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Notice that we can factor the left-hand side of the equation as follows: x 4 − 4 x 2 y + 3 x 2 y 2 + 5 x 2 − 1 2 y 3 + 1 5 y 2 = x 2 ( x 2 − 4 y ) + 3 x 2 y 2 − 1 2 y 3 + 5 x 2 + 1 5 y 2 = x 2 ( x 2 − 4 y ) + 3 y 2 ( x 2 − 4 y ) + 5 x 2 + 1 5 y 2 = ( x 2 + 3 y 2 ) ( x 2 − 4 y ) + 5 ( x 2 + 3 y 2 ) = ( x 2 + 3 y 2 ) ( x 2 − 4 y + 5 ) . Therefore, either x 2 + 3 y 2 = 0 or x 2 − 4 y + 5 = 0 . However, since x and y are positive, x 2 + 3 y 2 > 0 , so it must be the case that x 2 − 4 y + 5 = 0 . Therefore, 4 y − x 2 = 5 .
Note: I'd like to provide my thought process for the factoring above. To begin, I factored the first two terms just to see what happened, and bam! x 2 − 4 y shows up, which is the exact opposite of the quantity we want to find. We're making progress! Now I wanted to make more factors of x 2 − 4 y appear, and after a while then I found that 3 x 2 y 2 − 1 2 y 3 factors nicely, since the − 1 2 is − 4 times the 3 . After that, I grouped the two terms together and finished off the last two terms, giving that nice factorization.