Question

find the roots of the equation 1/2x-3 + 1/x-5=1 ;x not equal to 3/2,5 ​

Answer 1

$Given \: \frac{1}{(2x-3)} + \frac{1}{(x-5)} = 1$

$\implies \frac{x-5 + 2x - 3}{(2x-3)(x-5)}=1$

$\implies \frac{3x-8}{2x^{2}-10x-3x+15} = 1$

$\implies \frac{3x-8}{2x^{2}-13x+15} = 1$

$\implies 3x - 8 = (2x^{2} - 13x + 15 )$

$\implies 0 = 2x^{2} - 13x + 15 - 3x + 8$

$\implies 2x^{2} - 16x+ 23 = 0$

/* Compare above equation with ax² + bx + c = 0,we get */

$a = 2 , \: b = -16, \:c = 23$

$Discreminant (D) = b^{2} - 4ac \\= (-16)^{2} - 4\times 2\times 23\\= 256 - 184 \\= 72$

/* By Quadratic Formula: */

$x = \frac{-b\pm \sqrt{D}}{2a} \\= \frac{-(-16)\pm \sqrt{72}}{2\times 2}\\= \frac{16±6\sqrt{2}}{4}\\= \frac{2(8\pm 3\sqrt{2}}{4}\\= \frac{8\pm 3\sqrt{2}}{2}$

Therefore.,

$\red{ Roots \:of \: the \: equation } \green { = \frac{8\pm 3\sqrt{2}}{2}}$

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Answer 2

Step-by-step explanation:

Given

(2x−3)

1

+

(x−5)

1

=1

\implies \frac{x-5 + 2x - 3}{(2x-3)(x-5)}=1⟹

(2x−3)(x−5)

x−5+2x−3

=1

\implies \frac{3x-8}{2x^{2}-10x-3x+15} = 1⟹

2x

2

−10x−3x+15

3x−8

=1

\implies \frac{3x-8}{2x^{2}-13x+15} = 1⟹

2x

2

−13x+15

3x−8

=1

\implies 3x - 8 = (2x^{2} - 13x + 15 )⟹3x−8=(2x

2

−13x+15)

\implies 0 = 2x^{2} - 13x + 15 - 3x + 8⟹0=2x

2

−13x+15−3x+8

\implies 2x^{2} - 16x+ 23 = 0⟹2x

2

−16x+23=0

/* Compare above equation with ax² + bx + c = 0,we get */

a = 2 , \: b = -16, \:c = 23a=2,b=−16,c=23

\begin{lgathered}Discreminant (D) = b^{2} - 4ac \\= (-16)^{2} - 4\times 2\times 23\\= 256 - 184 \\= 72\end{lgathered}

Discreminant(D)=b

2

−4ac

=(−16)

2

−4×2×23

=256−184

=72

/* By Quadratic Formula: */

\begin{lgathered}x = \frac{-b\pm \sqrt{D}}{2a} \\= \frac{-(-16)\pm \sqrt{72}}{2\times 2}\\= \frac{16±6\sqrt{2}}{4}\\= \frac{2(8\pm 3\sqrt{2}}{4}\\= \frac{8\pm 3\sqrt{2}}{2}\end{lgathered}

x=

2a

−b±

D

=

2×2

−(−16)±

72

=

4

16±6

2

=

4

2(8±3

2

=

2

8±3

2

Therefore.,

\red{ Roots \:of \: the \: equation } \green { = \frac{8\pm 3\sqrt{2}}{2}}Rootsoftheequation=

2

8±3

2

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