Question
Find the roots
Find the roots of the algebra expression
x∈/R
Evaluate
x4−5x2+40
To find the roots of the expression,set the expression equal to 0
x4−5x2+40=0
Solve the equation using substitution t=x2
t2−5t+40=0
Substitute a=1,b=−5 and c=40 into the quadratic formula t=2a−b±b2−4ac
t=25±(−5)2−4×40
Simplify the expression
More Steps

Evaluate
(−5)2−4×40
Multiply the numbers
(−5)2−160
Rewrite the expression
52−160
Evaluate the power
25−160
Subtract the numbers
−135
t=25±−135
Simplify the radical expression
More Steps

Evaluate
−135
Evaluate the power
135×−1
Evaluate the power
135×i
Evaluate the power
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Evaluate
135
Write the expression as a product where the root of one of the factors can be evaluated
9×15
Write the number in exponential form with the base of 3
32×15
The root of a product is equal to the product of the roots of each factor
32×15
Reduce the index of the radical and exponent with 2
315
315×i
t=25±315×i
Separate the equation into 2 possible cases
t=25+315×it=25−315×i
Simplify the expression
t=25+2315it=25−315×i
Simplify the expression
t=25+2315it=25−2315i
Substitute back
x2=25+2315ix2=25−2315i
Solve the equation for x
More Steps

Substitute back
x2=25+2315i
Simplify
x=25+2315i
Rewrite the complex number in polar form
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Evaluate
25+2315i
Determine the modulus and the argument of the complex number
r=(25)2+(2315)2θ=arctan252315
Calculate
r=210θ=arctan252315
Substitute the given values into the formula r(cosθ+isinθ)
210×cosarctan252315+isinarctan252315
x=210×cosarctan252315+isinarctan252315
Calculate the nth roots of a complex r(cos(θ)+i×sin(θ),using nz=nr(cosnθ+2kπ+isinnθ+2kπ)
x=210×cos2arctan(252315)+2kπ+isin2arctan(252315)+2kπ
Simplify
x=440×cos2arctan(252315)+2kπ+isin2arctan(252315)+2kπ
Since n=2,substitute k=0,1 into the expression
x1=440×cos2arctan(252315)+2×0×π+isin2arctan(252315)+2×0×πx2=440×cos2arctan(252315)+2×1×π+isin2arctan(252315)+2×1×π
Calculate
More Steps

Evaluate
2arctan(252315)+2×0×π
Any expression multiplied by 0 equals 0
2arctan(252315)+0
Add the terms
2arctan(5315)
x1=440×cos2arctan(5315)+isin2arctan(5315)x2=440×cos2arctan(252315)+2×1×π+isin2arctan(252315)+2×1×π
Calculate
More Steps

Evaluate
2arctan(252315)+2×1×π
Multiply the terms
2arctan(252315)+2π
Evaluate the power
2arctan(5315)+2π
x1=440×cos2arctan(5315)+isin2arctan(5315)x2=440×cos2arctan(5315)+2π+isin2arctan(5315)+2π
Calculate
x1=440×cos2arctan(5315)+440×sin2arctan(5315)×ix2=440×cos2arctan(5315)+2π+440×sin2arctan(5315)+2π×i
x1=440×cos2arctan(5315)+440×sin2arctan(5315)×ix2=440×cos2arctan(5315)+2π+440×sin2arctan(5315)+2π×ix2=25−2315i
Solve the equation for x
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Substitute back
x2=25−2315i
Simplify
x=25−2315i
Rewrite the complex number in polar form
More Steps

Evaluate
25−2315i
Determine the modulus and the argument of the complex number
r=(25)2+(−2315)2θ=arctan25−2315
Calculate
r=210θ=arctan25−2315
Since 25−2315i lies in the IV quadrant, add 2π to get the argument in the IV quadrant
r=210θ=arctan−252315+2π
Simplify
r=210θ=arctan(−5315)+2π
Substitute the given values into the formula r(cosθ+isinθ)
210×(cos(arctan(−5315)+2π)+isin(arctan(−5315)+2π))
x=210×(cos(arctan(−5315)+2π)+isin(arctan(−5315)+2π))
Calculate the nth roots of a complex r(cos(θ)+i×sin(θ),using nz=nr(cosnθ+2kπ+isinnθ+2kπ)
x=210×cos2arctan(−5315)+2π+2kπ+isin2arctan(−5315)+2π+2kπ
Simplify
x=440×cos2arctan(−5315)+2π+2kπ+isin2arctan(−5315)+2π+2kπ
Since n=2,substitute k=0,1 into the expression
x1=440×cos2arctan(−5315)+2π+2×0×π+isin2arctan(−5315)+2π+2×0×πx2=440×cos2arctan(−5315)+2π+2×1×π+isin2arctan(−5315)+2π+2×1×π
Calculate
More Steps

Evaluate
2arctan(−5315)+2π+2×0×π
Any expression multiplied by 0 equals 0
2arctan(−5315)+2π+0
Removing 0 doesn't change the value,so remove it from the expression
2arctan(−5315)+2π
x1=440×cos2arctan(−5315)+2π+isin2arctan(−5315)+2πx2=440×cos2arctan(−5315)+2π+2×1×π+isin2arctan(−5315)+2π+2×1×π
Calculate
x1=440×cos2arctan(−5315)+2π+isin2arctan(−5315)+2πx2=440×cos2arctan(−5315)+2π+2π+isin2arctan(−5315)+2π+2π
Calculate
x1=−440×cos−2arctan(−5315)+440×sin2arctan(−5315)+2π×ix2=440×cos2arctan(−5315)+2π+2π+440×sin2arctan(−5315)+2π+2π×i
x1=440×cos2arctan(5315)+440×sin2arctan(5315)×ix2=440×cos2arctan(5315)+2π+440×sin2arctan(5315)+2π×ix1=−440×cos−2arctan(−5315)+440×sin2arctan(−5315)+2π×ix2=440×cos2arctan(−5315)+2π+2π+440×sin2arctan(−5315)+2π+2π×i
Solution
x∈/R
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