# Class 10 Math Sample Paper 2019 | Q23B

#### Quadratic Equations

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)]

## NCERT Solution to Class 10 Maths

### 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

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