If #f(x)=x^3-7x^2+5x#, how do you find all values for f(x)=-1?

3 Answers
Mar 31, 2018

#x_1=1#
#x_2=3+sqrt(10)~~6.162#
# x_3=3-sqrt(10)~~-0.162#

Explanation:

#-1=x^3-7x^2+5x|+1#
#0=x^3-7x^2+5x+1#
#0=(x^2-6x-1)(x-1)#

#x_1=1 or 0=x^2-6x-1#

#0=x^2-6x-1#
#0=(x-3)^2-9-1|+10#
#10=(x-3)^2|sqrt()|+3#
#3+-sqrt(10)=x_(2,3)#

#x_2=3+sqrt(10) or x_3=3-sqrt(10)#

Mar 31, 2018

Values are # x=1, x=3+ sqrt10 and x=3- sqrt10#
for #f(x)=-1#.

Explanation:

#f(x)= x^3-7x^2+5x ; f(x)= -1#

#:. x^3-7x^2+5x = -1# or

# x^3-7x^2+5x +1=0 # or

# x^3-x^2-6x^2+6x-x +1=0 # or

# x^2(x-1)-6x(x-1)-1(x-1)=0 # or

#(x-1)(x^2-6x-1)=0 :. (x-1)=0 or x=1# and

#x^2-6x-1=0 ; #Comparing with standard quadratic equation

#ax^2+bx+c=0; a=1 ,b=-6 ,c=-1# Discriminant

# D= b^2-4ac ; D=36+4=40#, discriminant is positive, we get

two real solutions, Quadratic formula: #x= (-b+-sqrtD)/(2a) #or

#x= (6+-sqrt(40))/2 or x= 3+- sqrt10#

#:. x=3+ sqrt10 and x=3- sqrt10 #

Values are # x=1, x=3+ sqrt10 and x=3- sqrt10#

for #f(x)=-1#. [Ans]

Mar 31, 2018

#x=3-sqrt10,1,3+sqrt10#

Explanation:

#x^3-7x^2+5x=-1#

#x^3-7x^2+5x+1=0#

By inspection,

Let

#g(x)=x^3-7x^2+5x+1#

For #x=1#

#g(1)=(1)^3-7(1)^2+5(1)+1#

#=1-7xx1+5+1#

#=1-7+5+1#

#=0#

#g(1)=0#

1 is a root

#x-1#
is a factor

Now,

#g(x)=h(x)(x-1)#

#x^2(x-1)=x^3-x^2#

#x^3-7x^2+5x+1=x^3-x^2-6x^2+5x+1#

#x^3-7x^2+5x+1=x^2(x-1)-6x^2+5x+1#

#-6x(x-1)=-6x^2+6x#

#x^2(x-1)-6x^2+5x+1=x^2(x-1)-6x^2+6x-x+1#

#x^2(x-1)-6x^2+5x+1=x^2(x-1)-6x(x-1)-x+1#

#x^3-7x^2+5x+1=x^2(x-1)-6x(x-1)-x+1#

#-x+1=-1(x-1)#

#x^2(x-1)-6x(x-1)-x+1=x^2(x-1)-6x(x-1)-1(x-1)#

#x^3-7x^2+5x+1=x^2(x-1)-6x(x-1)-1(x-1)#

Rearranging

#x^3-7x^2+5x+1=(x-1)(x^2-6x-1)#

By the method of completing the squares

#x^2-6x-1=(x^2-6x+9)-(9+1)#

#x^2-6x+9=(x-3)^2#

#-(9+1)=-10#

#x^2-6x-1=(x-3)^2-10#

#x^2-6x-1=(x-3)^2-(sqrt10)^2#

#(a^2-b^2=(a-b)(a+b)#

#(x-3)^2-(sqrt10)^2=((x-3)-sqrt10)((x-3)+sqrt10)#

Simplifying

#x^2-6x-1=(x-(3+sqrt10))(x-(3-sqrt10))#

#(x-1)(x^2-6x-1)=#
#(x-1)(x-(3+sqrt10))(x-(3-sqrt10))#

Thus,

#x^3-7x^2+5x+1=#
#(x-1)(x-(3+sqrt10))(x-(3-sqrt10))#

Thus, f(x)=-1, at

#x=1, 3+sqrt10, 3-sqrt10#

Arranging in the Increasing order

#x=3-sqrt10,1,3+sqrt10#