Question

Find the roots of the quadratic equation by factorisation.
x-1/x-2+x-3/x-4=10/3​

Answer 1

Answer:

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Answer 2

$\frac{x \: - \: 1}{x \: - \: 2} \: + \: \frac{x \: - \: 3}{x \: - \: 4} \: = \: \frac{10}{3}$

___________ [ GIVEN ]

• We have to find the value of x.

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→ $\frac{x \: - \: 1}{x \: - \: 2} \: + \: \frac{x \: - \: 3}{x \: - \: 4} \: = \: \frac{10}{3}$

→ $\frac{(x \: - \: 1)(x \: - \: 4) \: + \: (x \: - \: 3)(x \: - \: 2)}{(x \: - \: 2)(x \: - \: 4)} \: = \: \frac{10}{3}$

→ $\frac{ {x}^{2} \: - \: 4x \: - \: x \: + \: 4 \: + \: {x}^{2} \: 3x \: - \: 2x \: + \: 6 }{ {x}^{2} \: - \: 4x \: - \: 2x \: + \: 8 } \: = \: \frac{10}{3}$

→ $\frac{2 {x}^{2} \: - \: 10x \: + \: 10}{ {x}^{2} \: - \: 6x \: + \: 8 } \: = \: \frac{10}{3}$

Cross-multiply them

→ $3(2 {x}^{2} \: - \: 10x \: + \: 10) \: = \: 10({x}^{2} \: - \: 6x \: + \: 8)$

→ $6 {x}^{2} \: - \: 30x \: + \: 30 \: = \: 10{x}^{2} \: - \: 60x \: + \: 80$

→ $6 {x}^{2} \: - \: 10 {x}^{2} \: - \: 30x \: + \: 60x \: + \: 30 \: - \: 80\:=\:0$

→ $-\:4 {x}^{2} \: +\:30x \: - \: 50\:=\:0$

→ $4 {x}^{2} \: - \:30x \: + \: 50\:=\:0$

→ $2 {x}^{2} \: - \:15x \: + \: 25\:=\:0$

→ $2 {x}^{2} \: - \:10x \: - \: 5x + \: 25\:=\:0$

→ $2x(x\:-\:5) \: - \: 5(x - \: 5)\:=\:0$

→ $(2x \: - \: 5)\:(x \: - \: 5) \: = \: 0$

→ $x\:=\:\frac{5}{2}, \: +5$

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