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

Mar 31, 2018

${x}_{1} = 1$
${x}_{2} = 3 + \sqrt{10} \approx 6.162$
${x}_{3} = 3 - \sqrt{10} \approx - 0.162$

Explanation:

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

${x}_{1} = 1 \mathmr{and} 0 = {x}^{2} - 6 x - 1$

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

${x}_{2} = 3 + \sqrt{10} \mathmr{and} {x}_{3} = 3 - \sqrt{10}$

Mar 31, 2018

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

Explanation:

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

$\therefore {x}^{3} - 7 {x}^{2} + 5 x = - 1$ or

${x}^{3} - 7 {x}^{2} + 5 x + 1 = 0$ or

${x}^{3} - {x}^{2} - 6 {x}^{2} + 6 x - x + 1 = 0$ or

${x}^{2} \left(x - 1\right) - 6 x \left(x - 1\right) - 1 \left(x - 1\right) = 0$ or

$\left(x - 1\right) \left({x}^{2} - 6 x - 1\right) = 0 \therefore \left(x - 1\right) = 0 \mathmr{and} x = 1$ 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: $x = \frac{- b \pm \sqrt{D}}{2 a}$or

$x = \frac{6 \pm \sqrt{40}}{2} \mathmr{and} x = 3 \pm \sqrt{10}$

$\therefore x = 3 + \sqrt{10} \mathmr{and} x = 3 - \sqrt{10}$

Values are $x = 1 , x = 3 + \sqrt{10} \mathmr{and} x = 3 - \sqrt{10}$

for $f \left(x\right) = - 1$. [Ans]

Mar 31, 2018

$x = 3 - \sqrt{10} , 1 , 3 + \sqrt{10}$

Explanation:

${x}^{3} - 7 {x}^{2} + 5 x = - 1$

${x}^{3} - 7 {x}^{2} + 5 x + 1 = 0$

By inspection,

Let

$g \left(x\right) = {x}^{3} - 7 {x}^{2} + 5 x + 1$

For $x = 1$

$g \left(1\right) = {\left(1\right)}^{3} - 7 {\left(1\right)}^{2} + 5 \left(1\right) + 1$

$= 1 - 7 \times 1 + 5 + 1$

$= 1 - 7 + 5 + 1$

$= 0$

$g \left(1\right) = 0$

1 is a root

$x - 1$
is a factor

Now,

$g \left(x\right) = h \left(x\right) \left(x - 1\right)$

${x}^{2} \left(x - 1\right) = {x}^{3} - {x}^{2}$

${x}^{3} - 7 {x}^{2} + 5 x + 1 = {x}^{3} - {x}^{2} - 6 {x}^{2} + 5 x + 1$

${x}^{3} - 7 {x}^{2} + 5 x + 1 = {x}^{2} \left(x - 1\right) - 6 {x}^{2} + 5 x + 1$

$- 6 x \left(x - 1\right) = - 6 {x}^{2} + 6 x$

${x}^{2} \left(x - 1\right) - 6 {x}^{2} + 5 x + 1 = {x}^{2} \left(x - 1\right) - 6 {x}^{2} + 6 x - x + 1$

${x}^{2} \left(x - 1\right) - 6 {x}^{2} + 5 x + 1 = {x}^{2} \left(x - 1\right) - 6 x \left(x - 1\right) - x + 1$

${x}^{3} - 7 {x}^{2} + 5 x + 1 = {x}^{2} \left(x - 1\right) - 6 x \left(x - 1\right) - x + 1$

$- x + 1 = - 1 \left(x - 1\right)$

${x}^{2} \left(x - 1\right) - 6 x \left(x - 1\right) - x + 1 = {x}^{2} \left(x - 1\right) - 6 x \left(x - 1\right) - 1 \left(x - 1\right)$

${x}^{3} - 7 {x}^{2} + 5 x + 1 = {x}^{2} \left(x - 1\right) - 6 x \left(x - 1\right) - 1 \left(x - 1\right)$

Rearranging

${x}^{3} - 7 {x}^{2} + 5 x + 1 = \left(x - 1\right) \left({x}^{2} - 6 x - 1\right)$

By the method of completing the squares

${x}^{2} - 6 x - 1 = \left({x}^{2} - 6 x + 9\right) - \left(9 + 1\right)$

${x}^{2} - 6 x + 9 = {\left(x - 3\right)}^{2}$

$- \left(9 + 1\right) = - 10$

${x}^{2} - 6 x - 1 = {\left(x - 3\right)}^{2} - 10$

${x}^{2} - 6 x - 1 = {\left(x - 3\right)}^{2} - {\left(\sqrt{10}\right)}^{2}$

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

${\left(x - 3\right)}^{2} - {\left(\sqrt{10}\right)}^{2} = \left(\left(x - 3\right) - \sqrt{10}\right) \left(\left(x - 3\right) + \sqrt{10}\right)$

Simplifying

${x}^{2} - 6 x - 1 = \left(x - \left(3 + \sqrt{10}\right)\right) \left(x - \left(3 - \sqrt{10}\right)\right)$

$\left(x - 1\right) \left({x}^{2} - 6 x - 1\right) =$
$\left(x - 1\right) \left(x - \left(3 + \sqrt{10}\right)\right) \left(x - \left(3 - \sqrt{10}\right)\right)$

Thus,

${x}^{3} - 7 {x}^{2} + 5 x + 1 =$
$\left(x - 1\right) \left(x - \left(3 + \sqrt{10}\right)\right) \left(x - \left(3 - \sqrt{10}\right)\right)$

Thus, f(x)=-1, at

$x = 1 , 3 + \sqrt{10} , 3 - \sqrt{10}$

Arranging in the Increasing order

$x = 3 - \sqrt{10} , 1 , 3 + \sqrt{10}$