Question

find whether the equation 1/2x-3 +1/x-5=1 has real roots if real roots exist find them​

Answer 1

Step-by-step explanation:

By calculation we get

(X-5) + (2x-3) = (2x-3) (x-5)

3x-8 = 2x^2 - 10x - 3x +15

2x^2 - 16x +23 =0

Now check

b^2 - 4ac

256 - 184

27 > 0

So it has real roots

- b +sqrt b^2 - 4ac /2a

Or

- b - sqrt b^2 - 4ac /2a

16+ sqrt 27 / 8 और 16- sqrt 27 /8

Answer 2

Equation

$\rm \frac{1}{2x - 3} + \frac{1}{x - 5} = 1 \\ \rm \implies \frac{x - 5 + 2x - 3}{(2x - 3)(x - 5)} = 1 \\ \implies \rm\frac{3x - 8}{2 {x}^{2} - 10x - 3x + 15} = 1 \\ \rm \implies 3x - 8 = 2 {x}^{2} - 13x + 15 \\ \rm \implies2 {x}^{2} - 13x - 3x + 15 + 8 = 0\\ \implies \boxed{ \rm2 {x}^{2} - 16x + 23 = 0}$

Now we have Quadratic equations. Now we find discriminant

Discriminant

$\rm d = {b}^{2} - 4ac$

Here,

• a=2
• b=-16
• c=23

$\rm d = {( - 16)}^{2} - 4 \times 2 \times 23 \\ \implies \rm \: d = 256 - 184 \\ \implies \boxed{ \rm \:d = 72 }$

$\because \rm \: d > 0 \\ \therefore \rm roots \: are \: real$

Using Quadratic Formula

$\rm \: x = \frac{ - b \pm \sqrt{d} }{2a} \\ \rm \implies x = \frac{16 \pm \sqrt{72} }{2 \times 2} \\ \rm \implies \: x = \frac{16 \pm6 \sqrt{2} }{4} \\ \implies \rm \: x = \frac{2(8 \pm 3 \sqrt{2} )}{4} \\ \implies \boxed{ \rm \:x = \frac{8 \pm3 \sqrt{2} }{2} }$