What is the value of the constant k if the real solutions to the equation #x^4-kx^3+2kx^2+2x-20=0# are #x=2# and #x=-1#?

Question

1276 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-10-03T16:13:49

#k = 7#

Given: #x^4-kx^3+2kx^2+2x-20=0#

Substitute -1 for x:

#(-1)^4-k(-1)^3+2k(-1)^2+2(-1)-20=0#

#1 + k+2k -2 - 20 = 0#

#3k-21=0#

#k = 7#

Substitute 2 for x:

#(2)^4-k(2)^3+2k(2)^2+2(2)-20=0#

#16-8k+8k+ 4-20=0#

#0 = 0#

This means that all real values of k will give the polynomial a root of #x = 2#, therefore, well select the most restrictive, #k = 7#

Answer 2 · 2017-10-03T16:19:42

See below.

According to the question

#x^4 - k x^3 + 2 k x^2 + 2 x - 20 =(x - 2) (x + 1) (a x^2 + b x + c)#

or grouping coefficients

#{(2 c-20=0), (2 + 2 b + c=0), (2 a + b - c + 2 k=0), (a - b - k=0), (1 - a=0):}#

Solving for #a,b,c,k# we obtain

#a = 1, b = -6, c = 10, k = 7#