Question

find weather the following equation have real roots if real root exist find them 1/(2x-3)+1/(x-3)=1
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Answer 1

1/(2x-3)+1/(x-3)=1

x-3+2x-3/(2x-3)(x-3)=1

3x-6=(2x-3)(x-3)

3x-6=2x²-6x-3x+9

0=2x²-9x-3x+9+6

0=2x²-12x+15

here, a=2, b=-12, c=15

Discriminant value=b²-4ac

=>(-12)²-4(2)(15)

=>144-120

=>24>0

The equations has real and distinct roots.

By formula method:

x=-b±√b²-4ac/2a.

x=-(-12)±8√3/2(2).

x=12-8√3/4

x=3-2√3.

x=12+8√3/4

x=3+2√3.

3+2√3 and 3-2√3 are the roots of the equation 2x²-12x+15=0.