Class 10 Math Sample Paper 2019 | Q23B

This 2019 CBSE class 10 Maths 4 mark question is from Quadratic Equations. An medium difficulty 4 mark question in CBSE class 10 sample question paper.

Question 23B: Solve for x: \\frac{1}{(a + b + x)}) = \\frac{1}{a}) + \\frac{1}{b}) + \\frac{1}{x}) , [a ≠ 0, b ≠ 0, x ≠ 0, x ≠ -(a + b)]

Video Explanation

Explanatory Answer

\\frac{1}{(a + b + x)}) - \\frac{1}{x}) = \\frac{1}{a}) + \\frac{1}{b})
Take LCM of the denominators on both sides of the equation.
LCM(a + b + x, x) = (a + b + x)x ; LCM(a, b) = ab
\\frac{x}{(a + b + x)x}) - \\frac{(a + b + x)}{(a + b + x)x}) = \\frac{b}{ab}) + \\frac{a}{ab})
\\frac{− (a + b)}{(a + b + x)x}) = \\frac{(a + b)}{ab})
\\frac{− 1}{(a + b + x)x}) = \\frac{1}{ab})
Cross multiply the denominators on both sides of the equation.
-ab = x² + (a + b)x
x² + (a + b)x + ab = 0
x² + ax + bx + ab = 0
(x + a)(x + b) = 0
x = -a or x = -b