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

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

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Answer 1 · 2018-03-31T12:23:43

$x_{1} = 1$ $x_1=1$
$x_{2} = 3 + \sqrt{10} \approx 6.162$ $x_2=3+sqrt(10)~~6.162$
$x_{3} = 3 - \sqrt{10} \approx - 0.162$ $x_3=3-sqrt(10)~~-0.162$

Explanation:

$- 1 = x^{3} - 7 x^{2} + 5 x ∣ + 1$ $-1=x^3-7x^2+5x|+1$
$0 = x^{3} - 7 x^{2} + 5 x + 1$ $0=x^3-7x^2+5x+1$
$0 = (x^{2} - 6 x - 1) (x - 1)$ $0=(x^2-6x-1)(x-1)$

$x_{1} = 1 or 0 = x^{2} - 6 x - 1$ $x_1=1 or 0=x^2-6x-1$

$0 = x^{2} - 6 x - 1$ $0=x^2-6x-1$
$0 = {(x - 3)}^{2} - 9 - 1 ∣ + 10$ $0=(x-3)^2-9-1|+10$
$10 = {(x - 3)}^{2} ∣ ∣ \sqrt{} ∣ ∣ + 3$ $10=(x-3)^2|sqrt()|+3$
$3 \pm \sqrt{10} = x_{2, 3}$ $3+-sqrt(10)=x_(2,3)$

$x_{2} = 3 + \sqrt{10} or x_{3} = 3 - \sqrt{10}$ $x_2=3+sqrt(10) or x_3=3-sqrt(10)$

Answer 2 · 2018-03-31T12:40:47

Values are $x = 1, x = 3 + \sqrt{10} and x = 3 - \sqrt{10}$ $x=1, x=3+ sqrt10 and x=3- sqrt10$
for $f (x) = - 1$ $f(x)=-1$ .

Explanation:

$f (x) = x^{3} - 7 x^{2} + 5 x; f (x) = - 1$ $f(x)= x^3-7x^2+5x ; f(x)= -1$

$:. x^3-7x^2+5x = -1$ 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 · 2018-03-31T12:45:31

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

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

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

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

3 Answers

Explanation:

Explanation:

Explanation:

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