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student claims that the square root of -x equals three has no solution, since the square root of a negative number does not exist. Why is this argument wrong?
1 answer
There is a solution. The solution is an imaginary number.
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A student claims that the equation the !-x = 3 (square root of) has no solution, since the square root of a negative number does
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If you mean sqrt(-x) = 3 then as long as x is negative, you are taking the square root of a positive
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Which calculation can be used to find the value of q in the equation q3 = 64? (1 point)
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q equals cube root of 64
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B. Between square root 25 and square root 36
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A student claims that the equation �ã-x=3 has no solution, since the square root of a negative number does not exist.
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Use the quadratic formula to solve the equation. Give exact answers:
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My teacher said
If x-2(square root xy)=9-4(square root 2) so since there are two square roots -2(square root xy) equals -4(square
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your teacher is dead wrong. example: 6-sqrt25=15-sqrt196 sqrt25 is not equal to sqrt196, ever. Point
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A student claims that the equation√-x=3 has no solution, since the square root of a negative number does not exist. Why is
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What if x itself is negative ? e.g. √-(-9) = 3
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A student claims that the equation has no solution, since the square root of a negative number does not exist. Why is this
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Imaginary numbers.
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