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What is one value of x that makes the equation-16x^2+121=0 true

1 Answer

5 votes

Answer:

x = ±11/4

Explanation:

Rewrite -16x^2+121=0 as 16x^2 - 121=0 and then as 16x^2 = 121.

Take the square root of both sides. This returns 4x = ±11, and so:

x = ±11/4

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