Question

find C when f(x) = (x-4)(x-6)(x-8) on [4, 8]​

Answer 1

Step-by-step explanation:

\begin{gathered}\frac{x - 2}{x - 4} + \frac{x - 6}{x - 8} = \frac{20}{3} \\\end{gathered}

x−4

x−2

+

x−8

x−6

=

3

20

$$\textsc{On taking LCM we get}$$

$$\begin{gathered}\frac{(x - 2)(x - 8) + (x - 6)(x - 4)}{(x - 4)(x - 8)} = \frac{20}{3} \\ \\ \\ \\ \frac{ ({x}^{2} - 8x - 2x + 16) + ( {x}^{2} - 4x - 6x + 24)}{ {x}^{2} - 8x - 4x + 32} = \frac{20}{3} \\ \\ \frac{ ({x}^{2} - 10x + 16) + ( {x}^{2} - 10x + 24)}{ {x}^{2} - 12x + 32} = \frac{20}{3} \\ \\ \frac{ {x}^{2} - 10x + 16 + {x}^{2} - 10x + 24}{ {x}^{2} - 12x + 32} = \frac{20}{3} \\\end{gathered}$$

$$\textsc{Writing the like terms together}$$

$$\begin{gathered}\frac{ {x}^{2} + {x}^{2} - 10x - 10x + 16 + 24}{ {x}^{2} - 12x + 32 } = \frac{20}{3} \\ \\ \frac{2 {x}^{2} - 20x + 40}{ {x}^{2} - 12x + 32 } = \frac{20}{3} \\\end{gathered}$$

$$\textsc{On Cross Multiplying we get}$$

$$\begin{gathered}3(2 {x}^{2} - 20x + 40) = 20( {x}^{2} - 12x + 32) \\ \\ 6 {x}^{2} - 60x + 120 = 20 {x}^{2} - 240x + 640 \\ \\ 6 {x}^{2} - 20 {x}^{2} - 60x + 240x + 120 - 640 = 0 \\ \\ - 14 {x}^{2} + 180x - 520 = 0 \\ \\ - 2(7 {x}^{2} - 90x + 260) = 0 \\ \\ 7 {x}^{2} - 90x + 260 = 0 \\\end{gathered}$$

But since, we can't factorize it, let us use another formula i.e.

$$\boxed{\boxed{\mathbf{x = \frac{ - b ± \sqrt{b² - 4ac}}{2a}}}}$$

$$\begin{gathered}x = \frac{ - ( - 90) ± \sqrt{ { ( - 90) }^{2} - 4(7)(260) } }{2(7)} \\ \\ x = \frac{90 ± \sqrt{8100 - 7280} }{14} \\ \\ x = \frac{90 ± \sqrt{820} }{14} \\ \\ x = \frac{90 ± 2 \sqrt{205} }{14} \\ \\ x = \frac{2(45 ± \sqrt{205}) }{2(7)} \\ \\ x = \frac{45 ± \sqrt{205} }{7}\end{gathered}$$

So, The value of x is $$\frac{45 + \sqrt{205}}{7}$$ or $$\frac{45 - \sqrt{205}}{7}$$