a) 0 = x⁴ - 81
0 = (x² - 9)(x² + 9)
0 = (x - 3)(x + 3)(x² + 9)
0 = x - 3 , 0 = x + 3 , 0 = x² + 9
x = 3 , x = -3 , x² = -9 → x = √-9 → x = +/- 3i
Answer: real roots 3, 3 ; complex roots 3i, -3i
b) 0 = x⁴ + 10x² + 25
0 = (x² + 5)²
0 = x² + 5
x² = -5 → x = +/-√-5 → x = +/- i√5
Answer: no real roots ; complex roots i√5, -i√5
c) 0 = x⁴ - x² - 6
0 = (x² - 3)(x² + 2)
0 = x² - 3 0 = x² + 2
x² = 3 → x = +/-√3 x² = -2 → x = +/-√-2 → x = +/-i√2
Answer: real roots √3, -√3; complex roots i√2, -i√2
Answer:
A. x= 3 x=-3
B. No solution
C. x=[tex]\sqrt{3}[/tex]
Step-by-step explanation:
To answer the first one you just have to factorize into conjugate binomials.
[tex]x^{4}-81[/tex]= [tex](x^{2} +9)[/tex][tex](x^{2} -9)[/tex]
(x-3)(x+3)[tex](x^{2} +9)[/tex]
Now you put the binomials in the parenthesis and equal them to 0.
x-3=0
x=3
x+3=0
x=-3
[tex](x^{2} +9)[/tex]=0
[tex]x^{2}[/tex]=-9
x=[tex]\sqrt{-9}[/tex]
Since the square root of a negative number aint possible the only solutions for the equation are x=3 and x=-3
For the second one you just factorize it into two factors:
[tex]x^{4}+20x^{2}+25[/tex]= [tex](x^{2} +5)(x^{2} +5)[/tex]
Since both solutions end up in x=[tex]\sqrt{-5}[/tex] is an imaginary number and it has no solution.
The last one is also factorized into two binomials.
[tex]x^{4}-x^{2}-6[/tex]=[tex](x^{2} -3)(x^{2} +2)[/tex]
Now we put the content of the parenthesis into an equation.
[tex]x^{2} +2=0\\x^{2} -3=0[/tex]
Since the +2 will end up ina negative square root it is not possible.
So the only answer for this is x=[tex]\sqrt{3}[/tex]
