Can somebody help me with this one
2 answers:
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For a radical with an even root, like 2 in this case, if the radicand turns to a negative value, the result is just an imaginary value, which is another way to say, there's no such a root, no solution per se.
for that, let's first check when the radicand turns to 0 first.
so, that happens when x = 1, -8(1) +8, is just -8+8 or 0, ok.
so.. if "x" goes a bit higher, like say 2, -8(2) + 8 --> -16 + 8 --> -16
you'd get a negative value, and the radical doesn't have a root for that.
so... the domain, or values "x" can safely take without making the square root expression a negative value, are 1 or below, for example, if we use say -5
-8(-5) + 8 --> 40+8 --> 48 <------ a positive value
thus, (-∞, 1]
for that, let's first check when the radicand turns to 0 first.
so, that happens when x = 1, -8(1) +8, is just -8+8 or 0, ok.
so.. if "x" goes a bit higher, like say 2, -8(2) + 8 --> -16 + 8 --> -16
you'd get a negative value, and the radical doesn't have a root for that.
so... the domain, or values "x" can safely take without making the square root expression a negative value, are 1 or below, for example, if we use say -5
-8(-5) + 8 --> 40+8 --> 48 <------ a positive value
thus, (-∞, 1]