How to solve the equation 4x^4 + 5x^3 + 2x^2 + 5x + 4=0?
2 Answers
lmull | High School Teacher | (Level 1) Adjunct Educator
When you graph the equation, it hits -1 twice on the x axis; therefore, -1 is a root.
-1 / 4 5 2 5 4
4 1 1 4 0 use synthetic division
-1/ 4 1 1 4
4 -3 4 0 you get 4x^2-3x + 4 now use the quadratic equation
3 +- square root of 9 - 4(4)(4) all over 2(4)
3+- square root of -55 all over 8
Final answer: 3 +- i squareroot 55 all over 8
giorgiana1976 | College Teacher | (Level 3) Valedictorian
Being an even reciprocal equation, x=-1 is one if it's four roots.
We'll apply the following algorithm, to find out the other roots. The calculus will be made with the equation's coefficients:
(-1)*4+5=1
(-1)*1+2=1
(-1)*1+5=4
(-1)*4+4=0, when the final result is 0, that certifies "-1" as root.
We've obtained an equation of third degree:
4x^3+x^2+x+4=0
We'll apply again the same algorithm:
(-1)*4+1=-3
(-1)*-3+1=4
(-1)*4+4=0, the second root will be again "-1"
The equation obtained with Horner's algorithm is:
4x^2-3x+4=0
In order to find out the following 2 roots, we'll calculate delta:
delta=(-3)^2-4*4*4=9-64=55
x3=(-(-3)+sqrt 55)/2*4=(3+sqrt 55)/8
x4=(3-sqrt 55)/8