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AVprozaik [17]
3 years ago
14

Solve by factoring: x^4-12x^2= 64

Mathematics
2 answers:
USPshnik [31]3 years ago
7 0
The final solution is all the values that make <span><span><span><span><span>(<span>x+4</span>)</span><span>(<span>x<span>−4</span></span>)</span></span><span>(<span><span>x2</span>+4</span>)</span></span>=0</span><span><span><span><span>x+4</span>⁢<span>x<span>-4</span></span></span>⁢<span><span>x2</span>+4</span></span>=0</span></span> true.<span>x=<span>−4</span>,4,<span>2i</span>,<span><span>−2</span><span>i</span></span></span>
Mariana [72]3 years ago
3 0
Minus 64 from both sides
x⁴-12x²-64=0
hmm
what 2 numbers multiply to -64 and add to get -12
-16 and 4
(x²-16)(x²+4)=0
oh look a difference of 2 perfect squares
(x²-4²)(x²+4)=0
(x-4)(x+4)(x²+4)=0
set each to zero

x-4=0
x=4

x+4=0
x=-4

x²+4=0
x²=-4
if you have learend complex roots then
sqrt both sides to get
x=-2i or 2i



solutions are
x=-4, 4, 2i or -2i
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