Question

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

Answer 1

(- 3, 0), (2 +
`√(2)`, 0), (2 -
`√(2)`, 0)

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 ±
`√(16-(4(1)(2))`)/ 2

= (4 ±
`√(8)`) / 2

= (4 ± 2
`√(2)`) / 2

= 2 ±
`√(2)`, hence

x = 2 +
`√(2)`, x = 2 -
`√(2)`

intersect at (- 3, 0), (2 +
`√(2)`, 0), (2 -
`√(2)`, 0)

Answer 2

Answer:
`\bold{x=-3,\quad x=2+√(2), \quad x=2-√(2) }}`

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=(-b\pm√(b^2-4ac))/(2a)`

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

`x=(-(-4)\pm√((-4)^2-4(1)(2)))/(2(1))`

`=(4\pm√(16-8))/(2)`

`=(4\pm√(8))/(2)`

`=(4\pm2√(2))/(2)`

`=2\pm√(2)`

`x=2+√(2)\qquad x=2-√(2)`

Intercepts are
`x=-3,\quad x=2+√(2), \quad x=2-√(2)}`