The given equation involves a power of the variable. Find all real solutions of

aortiH

aortiH

Answered question

2021-09-22

The given equation involves a power of the variable. Find all real solutions of the equation.
x416=0

Answer & Explanation

Sally Cresswell

Sally Cresswell

Skilled2021-09-23Added 91 answers

Step 1
To find the real solutions of equation: x416=0
Solution:
x416=0
On simplifying further, we get:
x416=0
(x2)242=0
(x2+4)(x24)=0(using,(a2b2)=(a+b)(ab))
(x2+4)(x222)=0
(x2+4)(x+2)(x2)=0
(x2+4)=0,or,(x+2)=0,or,(x2)=0
considering, x2+4=0
we know, x20 for all xR
x2+4>0, for all xR
hence, no real solution from here.
considering, x+2=0, x=2.
considering, x2=0,x=2.
Hence, two real solutions: x=2,2
Step 2
Result:
x=2,2

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