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Find the number of ordered pairs (x,y) of real numbers such that (2x^2 + 4x + 2)(3y^2 - 6y + 3) = 0.

 Jun 19, 2024

Best Answer 

 #1
avatar+1252 
+1

We can split this equation into 2 equations. Since if either them equal 0 the equation will be true, we just set both quadratics to equal 0. 

We have

\(2x^2+4x+2=0\quad \mathrm{or}\quad \:3y^2-6y+3=0\)

 

Solving the first equation, we have

\(2(x^2+2x+1)=0\\ 2(x+1)^2=0\\ x=-1\)

 

Solving the second equation, we have

\(3(y^2+-2y+1)=0\\ 3(y-1)^2=0\\ y=1\)

 

So there are an infinite amount of ordered pairs for x and y as long as either \(x=-1\) or \(y=1\)

So infinite is our answer...

 

Thanks! :)

 Jun 19, 2024
 #1
avatar+1252 
+1
Best Answer

We can split this equation into 2 equations. Since if either them equal 0 the equation will be true, we just set both quadratics to equal 0. 

We have

\(2x^2+4x+2=0\quad \mathrm{or}\quad \:3y^2-6y+3=0\)

 

Solving the first equation, we have

\(2(x^2+2x+1)=0\\ 2(x+1)^2=0\\ x=-1\)

 

Solving the second equation, we have

\(3(y^2+-2y+1)=0\\ 3(y-1)^2=0\\ y=1\)

 

So there are an infinite amount of ordered pairs for x and y as long as either \(x=-1\) or \(y=1\)

So infinite is our answer...

 

Thanks! :)

NotThatSmart Jun 19, 2024

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