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Write an equation with integer solutions of:

x = ±6, y = 0
and
x = 0, y = ±12

A)x2 + y2 + 6 = 0
B)4x2 + y2 + 144 = 0
C)x2 + y2 − 6 = 0
D)4x2 + y2 − 1

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The correct answer is:

D. 4x2 + y2 − 1

[tex]|Huntrw6|[/tex]

Answer:  Option 'D' is correct.

Step-by-step explanation:

Since we have given that

x = ±6, y = 0

and x = 0 and y = ±12

And we need an equation which has above as an integer solutions.

So, it becomes,

Consider [tex]4x^2+y^2-144=0[/tex]

Put x = 0, we get

[tex]0+y^2=144\\\\y^2=144\\\\y=\sqrt{144}\\\\y=\pm 12[/tex]

Similarly,

Put y = 0,

[tex]4x^2+0=144\\\\4x^2=144\\\\x^2=\dfrac{144}{4}\\\\x^2=36\\\\\x=\pm 6[/tex]

Hence, Option 'D' is correct.

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