How to start this

Algebra Level 3

For some positive real numbers x x and y y , x 4 4 x 2 y + 3 x 2 y 2 + 5 x 2 12 y 3 + 15 y 2 = 0. x^4-4x^2y+3x^2y^2+5x^2-12y^3+15y^2=0. What is the value of 4 y x 2 4y-x^2 ?


The answer is 5.

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

20 solutions

Michael Tang
Sep 16, 2013

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 12 y 3 + 15 y 2 = x 2 ( x 2 4 y ) + 3 x 2 y 2 12 y 3 + 5 x 2 + 15 y 2 = x 2 ( x 2 4 y ) + 3 y 2 ( x 2 4 y ) + 5 x 2 + 15 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 ) . \begin{aligned} &\quad x^4 - 4x^2y + 3x^2y^2 + 5x^2 - 12y^3 + 15y^2 \\ &= x^2(x^2-4y) + 3x^2y^2 - 12y^3 + 5x^2 + 15y^2 \\ &= x^2(x^2-4y) + 3y^2(x^2-4y) + 5x^2 + 15y^2 \\ &= (x^2+3y^2)(x^2-4y) + 5(x^2+3y^2) \\ &= (x^2+3y^2)(x^2-4y+5). \end{aligned} Therefore, either x 2 + 3 y 2 = 0 x^2+3y^2 = 0 or x 2 4 y + 5 = 0. x^2-4y +5 = 0. However, since x x and y y are positive, x 2 + 3 y 2 > 0 , x^2+3y^2 > 0, so it must be the case that x 2 4 y + 5 = 0. x^2-4y +5 = 0. Therefore, 4 y x 2 = 5 . 4y - x^2 = \boxed{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 x^2-4y 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 x^2-4y appear, and after a while then I found that 3 x 2 y 2 12 y 3 3x^2y^2-12y^3 factors nicely, since the 12 -12 is 4 -4 times the 3. 3. After that, I grouped the two terms together and finished off the last two terms, giving that nice factorization.

hi MSTang!

Jeremy Lu - 7 years, 8 months ago

Log in to reply

Oh hi jeremylu!

Michael Tang - 7 years, 8 months ago

where did the exponent 4 go?

HanaNhor Aliman - 7 years, 8 months ago

Log in to reply

What do you mean exactly?

Michael Tang - 7 years, 8 months ago
Kunal Singh
Sep 15, 2013

The given equation can be factorized as follows : x 4 4 x 2 y + 3 x 2 y 2 + 5 x 2 12 y 3 + 15 y 2 x^4-4x^2y+3x^2y^2+5x^2-12y^3+15y^2

= ( x 4 4 x 2 y ) + ( 3 x 2 y 2 12 y 3 ) + ( 5 x 2 + 15 y 2 ) =(x^4-4x^2y)+(3x^2y^2-12y^3)+(5x^2+15y^2)

= x 2 ( x 2 4 y ) + 3 y 2 ( x 2 4 y ) + 5 ( x 2 + 3 y 2 ) =x^2(x^2-4y)+3y^2(x^2-4y)+5(x^2+3y^2)

= ( x 2 + 3 y 2 ) ( x 2 4 y ) + 5 ( x 2 + 3 y 2 ) =(x^2+3y^2)(x^2-4y)+5(x^2+3y^2)

= ( x 2 + 3 y 2 ) ( x 2 4 y + 5 ) =(x^2+3y^2)(x^2-4y+5)

Since the given equation equals zero , we have

( x 2 + 3 y 2 ) ( x 2 4 y + 5 ) = 0 (x^2+3y^2)(x^2-4y+5)=0 which means either x 2 + 3 y 2 = 0 x^2+3y^2=0 or x 2 4 y + 5 = 0 x^2-4y+5=0 . Considering the second condition , we get x 2 4 y + 5 = 0 4 y x 2 = 5 x^2-4y+5=0 \Rightarrow 4y-x^2=\boxed{5}

Chris Catacata
Sep 15, 2013

0 = ( x 4 4 x 2 y + 5 x 2 ) + ( 3 x 2 y 2 12 y 3 + 15 y 2 ) 0 =(x^4-4x^2y+5x^2)+(3x^2y^2-12y^3+15y^2)

0 = x 2 ( x 2 4 y + 5 ) + 3 y 2 ( x 2 4 y + 5 ) 0 =x^2(x^2-4y+5)+3y^2(x^2-4y+5)

0 = ( x 2 + 3 y 2 ) ( x 2 4 y + 5 ) 0 =(x^2+3y^2)(x^2-4y+5)

x 2 4 y + 5 = 0 x^2-4y+5=0

4 y x 2 = 5 4y-x^2=\fbox{5}

Moderator note:

You need to explain why we must have x 2 4 y + 5 = 0 x^2 - 4y + 5 = 0 . Why can't x 2 + 3 y 2 = 0 x^2 + 3y^2 = 0 ?

thanks..

Samuel Mondido - 7 years, 8 months ago
Tim Vermeulen
Sep 16, 2013

Note that

x 4 4 x 2 y + 3 x 2 y 2 + 5 x 2 12 y 3 + 15 y 2 = ( x 2 4 y + 5 ) ( x 2 + 3 y 2 ) = 0. \begin{aligned} x^4 - 4x^2y + 3x^2y^2 + 5x^2 - 12y^3 + 15y^2 &=\\ (x^2-4y+5)(x^2+3y^2) &= 0. \end{aligned}

We know that x x and y y are strictly positive, so

x 2 + 3 y 2 > 0 x 2 4 y + 5 = 0 , x^2+3y^2 > 0 \implies x^2 - 4y + 5 = 0,

hence

4 y x 2 = 5 . 4y-x^2 = \boxed{5}.

Geoffrey Mooney
Sep 16, 2013

x 4 4 x 2 y + 3 x 2 y 2 + 5 x 2 12 y 3 + 15 y 2 = 0 x^{4}-4x^{2}y+3x^{2}y^{2}+5x^{2}-12y^{3}+15y^{2}=0

x 4 + 3 x 2 y 2 + 5 x 2 4 x 2 y + 15 y 2 12 y 3 = 0 x^{4}+3x^{2}y^{2}+5x^{2}-4x^{2}y+15y^{2}-12y^{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}+3y^{2})+x^{2}(5-4y)+3y^{2}(5-4y)=0

x 2 ( x 2 + 3 y 2 ) + ( 5 4 y ) ( x 2 + 3 y 2 ) = 0 x^{2}(x^{2}+3y^{2})+(5-4y)(x^{2}+3y^{2})=0

( x 2 + 3 y 2 ) ( x 2 4 y + 5 ) = 0 (x^{2}+3y^{2})(x^{2}-4y+5)=0

x 2 4 y + 5 = 0 x^{2}-4y+5=0

4 y x 2 = 5 4y-x^{2}=5

Fengyu Seah
Sep 17, 2013

The goal will be to factorize the expression into something with 4 y x 2 4y - x^2 as a factor.

Therefore, x 4 4 x 2 y + 3 x 2 y 2 + 5 x 2 12 y 3 + 15 y 2 x^4 - 4x^2 y + 3x^2 y^2 + 5x^2 - 12y^3 + 15y^2 = x 2 ( 4 y x 2 ) 3 y 2 ( 4 y x 2 ) + 5 ( x 2 + 3 y 2 ) = -x^2 (4y - x^2) - 3y^2 (4y - x^2) + 5(x^2 + 3y^2) = ( 4 y x 2 ) ( x 2 + 3 y 2 ) + 5 ( x 2 + 3 y 2 ) = -(4y - x^2)(x^2 + 3y^2) + 5(x^2 + 3y^2) = ( x 2 4 y + 5 ) ( x 2 + 3 y 2 ) = 0 = (x^2 - 4y + 5)(x^2 + 3y^2) = 0

Since x x and y y are positive real numbers, both x 2 x^2 and 3 y 2 3y^2 will be positive. Thus, for the entire expression to be equals to 0 0 , x 2 4 y + 5 = 0 x^2 - 4y + 5 = 0 . Therefore, x 2 4 y = 5 x^2 - 4y = -5 , and 4 y x 2 = 5 4y - x^2 = 5

Visal In
Sep 16, 2013

x 4 4 x 2 y + 3 x 2 y 2 + 5 x 2 12 y 3 + 15 y 2 = 0 x^4 - 4x^2y + 3x^2y^2 + 5x^2 - 12y^3 + 15y^2 = 0

can reduce to

( x 2 + 3 y 2 ) ( 4 y x 2 5 ) (x^2 + 3y^2)(4y - x^2 - 5) = 0

Therefore

So 4 y x 2 = 5 4y - x^2 = 5 .

Moderator note:

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 x^2 + 3y^2 \neq 0 .

Jayver De Torres
Sep 16, 2013

Rearranging the equation, you can have x 4 + 3 x 2 y 2 4 x 2 y 12 y 3 + 5 x 2 + 15 y 2 = 0 x^4 + 3x^2y^2- 4x^2y- 12y^3 + 5x^2 +15y^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 x^2(x^2 +3y^2) -4y(x^2 +3y^2 ) +5(x^2 + 3y^2)= 0 . Hence, ( x 2 4 y + 5 ) ( x 2 + 3 y 2 ) = 0 (x^2 - 4y + 5)(x^2 +3y^2)= 0 . x 2 4 y + 5 = 0 x^2 - 4y +5 = 0 . x 2 4 y + = 5 x^2 - 4y + = -5 . 4 y x 2 = 5 4y - x^2 = 5

David Nolasco
Sep 16, 2013

Factoring the given equation will result to:

( x 2 + 3 y 2 ) ( x 2 4 y + 5 ) = 0 (x^{2} + 3y^{2})(x^{2} - 4y + 5) = 0 .

Note that since x x and y y are both positive real numbers, the first factor does not make the product zero.

Therefore x 2 4 y + 5 = 0 x^2 - 4y + 5 =0 . Rearranging the equation will give us 4 y x 2 = 5 4y - x^{2} =5

Therefore the answer is 5 .

Toan Pham Quang
Sep 15, 2013

The equation is equivalent to ( x 2 + 3 y 2 ) ( 5 4 y + x 2 ) = 0 (x^2+3y^2)(5-4y+x^2)=0 . Since $x,y>0$ then 5 4 y + x 2 = 0 5-4y+x^2=0 or x 2 4 y = 5 x^2-4y= \boxed{5} .

Ivan Sekovanić
Sep 17, 2013

Let us modify the expression a bit:

x 4 4 x 2 y + 3 x 2 y 2 + 5 x 2 12 y 3 + 15 y 2 = 0 x^{4}-4x^{2}y+3x^{2}y^{2}+5x^{2}-12y^{3}+15y^{2}=0 \Rightarrow

x 2 ( 4 y x 2 ) 3 y 2 ( 4 y x 2 ) + 5 ( x 2 + 3 y 2 ) = 0 \Rightarrow -x^{2}(4y-x^{2})-3y^{2}(4y-x^{2})+5(x^{2}+3y^{2})=0 \Rightarrow

( x 2 + 3 y 2 ) ( 4 y x 2 ) + 5 ( x 2 + 3 y 2 ) = 0 \Rightarrow -(x^{2}+3y^{2})(4y-x^{2})+5(x^{2}+3y^{2})=0 \Rightarrow

( x 2 + 3 y 2 ) ( 4 y x 2 ) = 5 ( x 2 + 3 y 2 ) \Rightarrow -(x^{2}+3y^{2})(4y-x^{2})=-5(x^{2}+3y^{2}) \Rightarrow

4 y x 2 = 5 ( x 2 + 3 y 2 ) ( x 2 + 3 y 2 ) \Rightarrow 4y-x^{2}=\frac{-5(x^{2}+3y^{2})}{-(x^{2}+3y^{2})} \Rightarrow

4 y x 2 = 5 \Rightarrow 4y-x^{2}=5

From this, we can clearly see that 4 y x 2 4y-x^{2} equals exactly 5 5 , which is the correct answer.

Kumar Ashutosh
Sep 17, 2013

Here we need to find the value of 4 y x 2 4y-x^2 , So let 4 y x 2 = k x 2 = 4 y k 4y-x^2=k \Rightarrow x^2=4y-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 ) 12 y 3 + 15 y 2 = 0 (4y-k)^2-4(4y-k)y+3(4y-k)^2y^2+5(4y-k)-12y^3+15y^2=0 which on simplifying reduces to k 2 4 y k 3 y 2 k + 20 y 5 k + 15 y 2 = 0 k^2-4yk-3y^2k+20y-5k+15y^2=0 3 ( 5 k ) y 2 + 4 ( 5 k ) y + k ( k 5 ) = 0 \Rightarrow 3(5-k)y^2+4(5-k)y+k(k-5)=0 ( 5 k ) ( 3 y 2 + 4 y k ) = 0 \Rightarrow (5-k)(3y^2+4y-k)=0 . Therefore we get k = 5 k=5 . Hence 4y-x^2=5

Brij Rokad
Sep 22, 2013

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

Samuel Hatin
Sep 21, 2013

This problem can be solved with factorization:

x 4 4 x 2 y + 3 x 2 y 2 + 5 x 2 12 y 3 + 15 y 2 = 0 3 x 2 y 2 + 5 x 2 12 y 3 + 15 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 + 5 ( x 2 + 3 y 2 ) ( x 2 4 y ) = x 2 5 ( x 2 + 3 y 2 ) ( x 2 4 y ) = ( x 2 + 3 y 2 ) 5 = ( x 2 4 y ) 5 = 4 y x 2 \begin{array}{l} {x^4} - 4{x^2}y + 3{x^2}{y^2} + 5{x^2} - 12{y^3} + 15{y^2} = 0\\ 3{x^2}{y^2} + 5{x^2} - 12{y^3} + 15{y^2} = {x^2}\left( {4y - {x^2}} \right)\\ 3{y^2}\left( {{x^2} - 4y} \right) + 5\left( {{x^2} + 3{y^2}} \right) = {x^2}\left( {4y - {x^2}} \right)\\ 3{y^2} + \frac{{5\left( {{x^2} + 3{y^2}} \right)}}{{\left( {{x^2} - 4y} \right)}} = - {x^2}\\ \frac{{5\left( {{x^2} + 3{y^2}} \right)}}{{\left( {{x^2} - 4y} \right)}} = - \left( {{x^2} + 3{y^2}} \right)\\ 5 = - \left( {{x^2} - 4y} \right)\\ 5 = 4y - {x^2} \end{array}

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

Anirudh Chauhan
Sep 21, 2013

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

Shivam Gulati
Sep 20, 2013

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.

Phúc Nguyễn
Sep 19, 2013

We rewrite the equation as ( x 2 4 y + 5 ) ( x 2 + 3 y 2 ) = 0 (x^2-4y+5)(x^2+3y^2)=0 and therefore x 2 4 y + 5 = 0 x^2-4y+5=0 or 4 y x 2 = 5 4y-x^2=5

Francis Naldo
Sep 19, 2013

First, simplify or factor the equation..

( x 4 x^{4} + 3 x 2 x^{2} y 2 y^{2} ) + ( 5 x 2 x^{2} + 15 y 2 y^{2} ) - ( 4 x 2 x^{2} y - 12 y 3 y^{3} ) = 0

then remove the common factor;

x 2 x^{2} ( x 2 x^{2} + 3 y 2 y^{2} ) + 5( x 2 x^{2} + 3 y 2 y^{2} ) - 4y( x 2 x^{2} + 3 y 2 y^{2} = 0

and since the 3 has common factor, we can unite them using distributive property;

( x 2 x^{2} - 4y + 5)( x 2 x^{2} + 3 y 2 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} );

            ( \\\( x^\{2\} \\\) - 4y + 5)
=>        4y - \\\( x^\{2\} \\\) = 5
Tristan Shin
Sep 18, 2013

By the Commutative Property, this equation also means x 4 4 x 2 y + 3 x 2 y 2 12 y 3 + 5 x 2 + 15 y 2 = 0 x^{4}-4x^{2}y+3x^{2}y^{2}-12y^{3}+5x^{2}+15y^{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 x^{2}(x^{2}-4y)+3y^{2}(x^{2}-4y)+5(x^{2}+3y^{2}) = (x^{2}+3y^{2})(x^{2}-4y)+5(x^{2}+3y^{2}) = 0 . Therefore, ( x 2 + 3 y 2 ) ( x 2 4 y ) = 5 ( x 2 + 3 y 2 ) (x^{2}+3y^{2})(x^{2}-4y) = -5(x^{2}+3y^{2}) , and x 2 4 y = 5 x^{2}-4y = -5 . Finally, 4 y x 2 = 5 4y-x^{2} = 5 .

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...