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# Algebra

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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

#1
+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 $$y=1$$

Thanks! :)

Jun 19, 2024

#1
+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 $$y=1$$