Question

What are the x–coordinates of the three points where the graphs of the equations intersect? x3 – 10x = x2 – 6

Answer 1

Answer:

(- 3, 0), (2 + $\sqrt{2}$, 0), (2 - $\sqrt{2}$, 0)

Step-by-step explanation:

rearrange the equation equating to zero

x³ - x² - 10x + 6 = 0

note that when x = - 3

(- 3)³ - (- 3)² - 10(- 3) + 6 = - 27 - 9 + 30 + 6 = 0, hence

x = - 3 is a root and (x + 3) is a factor of the polynomial

dividing x³ - x² - 10x + 6 ÷ (x + 3) gives

(x + 3)(x² - 4x + 2) = 0

solve x² - 4x + 2 = 0 using the quadratic formula

with a = 1, b= - 4 and c = 2

x = (4 ± $\sqrt{16-(4(1)(2)}$)/ 2

  = (4 ± $\sqrt{8}$) / 2

  = (4 ± 2$\sqrt{2}$) / 2

  = 2 ±$\sqrt{2}$, hence

x = 2 + $\sqrt{2}$, x = 2 - $\sqrt{2}$

intersect at (- 3, 0), (2 +$\sqrt{2}$, 0), (2 - $\sqrt{2}$, 0)

Answer 2

Answer: $\bold{x=-3,\quad x=2+\sqrt{2}, \quad x=2-\sqrt{2} }}$

Step-by-step explanation:

First, move all terms to one side with zero on the other side so the Zero Product Property can be applied:

x³ - x² - 10x + 6 = 0    Possible rational roots are: ±{1, 2, 3, 6}

Use synthetic division to find one of the roots.  Let's try x = -3

-3 |  1    -1    -10    6

   |  ↓   -3    12    -6

      1    -4     2     0  ← remainder is zero, so x = -3 is a root/intercept

Next, evaluate the reduced polynomial x² - 4x + 2 = 0

This can not be factored so use the quadratic formula to find the remaining roots/intercepts: $x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$

a = 1, b = -4, c = 2

$x=\dfrac{-(-4)\pm\sqrt{(-4)^2-4(1)(2)}}{2(1)}$

 $=\dfrac{4\pm\sqrt{16-8}}{2}$

 $=\dfrac{4\pm\sqrt{8}}{2}$

 $=\dfrac{4\pm2\sqrt{2}}{2}$

 $=2\pm\sqrt{2}$

$x=2+\sqrt{2}\qquad x=2-\sqrt{2}$

Intercepts are $x=-3,\quad x=2+\sqrt{2}, \quad x=2-\sqrt{2}}$