1/(2x - 3) + 1/(x-5) = 1, x ≠ 3/2, 5, find whether the equation has real roots. If real roots exist, find them
Solution:
Given, the equation is 1/(2x-3) + 1/(x-5) = 1
We have to find whether the equation has real roots or not.
The equation can be rewritten as
(x - 5) + (2x - 3) = (x -5)(2x - 3)
x - 5 + 2x - 3 = 2x² - 3x - 10x + 15
3x - 8 = 2x² - 13x + 15
By grouping,
2x² - 13x - 3x + 15 + 8 = 0
2x² - 16x + 23 = 0
A quadratic equation ax² + bx + c = 0 has 2 distinct real roots when the discriminant of the equation is greater than zero.
Discriminant = b² - 4ac
Here, a = 2, b = -16 and c = 23
So, b² - 4ac = (-16)² - 4(2)(23)
= 256 - 184
= 72 > 0
So, the equation has 2 distinct real roots.
By using the quadratic formula,
x = [-b ± √b² - 4ac]/2a
x = (16 ± √72)/2(2)
= (16 ± 6√2)/4
Now, x = (16 + 6√2)/4 = 2(8+3√2)/4 = (8+3√2)/2 = 2(4+3√2/2)/2 = 4+3√2
x = (2 - 6√2)/4 = 2(1-3√2)/4 = (1-3√2)/2 = 2(4-3√2/2)/2 = 4-3√2
Therefore, the roots of the equation are (4+3√2/2) and (4-3√2/2)
✦ Try This: Check whether the equation 2x² + 2x - 1 = 0 has real roots. If real roots exist, find them
☛ Also Check: NCERT Solutions for Class 10 Maths Chapter 4
NCERT Exemplar Class 10 Maths Exercise 4.4 Problem 1 (iv)
1/(2x - 3) + 1/(x-5) = 1, x ≠ 3/2, 5, find whether the equation has real roots. If real roots exist, find them
Summary:
The equation 1/(2x - 3) + 1/(x-5) = 1, x ≠ 3/2, 5, has real roots x = (4-3√2/2) and x = (4+3√2/2)
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