Question

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

Answer 1

`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)`

Answer 2

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]

Answer 3

`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`