The given 4th order polynomial equation only has real solutions, x = √7 and x = –√7
The given equation is x⁴-14 x²+49 = 0.
This is a 4th degree polynomial equation.
Let p = x².
Then we can write x⁴-14 x²+49 = 0 as:
p²- 14p +49 = 0
This is a quadratic equation.
There are three ways to find the solutions of a quadratic equation:
Notice that p²- 14p +49 = 0 can be written as a perfect square:
(p - 7)² = 0
Hence, it has only single real solution:
p - 7 = 0
p = 7
Substitute p = 7 into p = x², we get:
x² = 7
x = ± √7
The solutions are, therefore, x = √7 and x = –√7
Learn more about the solutions of polynomial equations here:
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