Solve- 1-18-x-1-18-x-1-24 and x-0-

(i) Given quadratic equation is 1/18 x 1/18+x=1/24 ⇒(18+x) (18 x)/(18+x)(18 x)=1/24 ⇒2 x/18^2 x^2=1/24 ⇒ 48 x=324 x^2 ⇒ x^2+48 x 324=0 ⇒ x^2+54 x 6 x 324=0 ⇒ x( ...

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Standard XIII

Mathematics

Roots of a Quadratic Equation

Solve: 1/18-x...

Question

Solve: $\frac{1}{18 - x} - \frac{1}{18 + x} = \frac{1}{24}$ and $x > 0$ .

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Solution

(i) Given quadratic equation is $\frac{1}{18 - x} - \frac{1}{18 + x} = \frac{1}{24}$ $\Rightarrow \frac{(18 + x) - (18 - x)}{(18 + x) (18 - x)} = \frac{1}{24}$ $\Rightarrow \frac{2 x}{18^{2} - x^{2}} = \frac{1}{24}$
$\Rightarrow 48 x = 324 - x^{2}$
$\Rightarrow x^{2} + 48 x - 324 = 0$
$\Rightarrow x^{2} + 54 x - 6 x - 324 = 0$
$\Rightarrow x (x + 54) - 6 (x + 54) = 0$
$\Rightarrow (x + 54) (x - 6) = 0$
$\Rightarrow x = - 54$ or $x = 6$
But as $x > 0$ , so $x$ can't be negative.
Hence, $x = 6$ .

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Solve: 118−x−118+x=124 and x>0.

(i) Given quadratic equation is 118−x−118+x=124 ⇒(18+x)−(18−x)(18+x)(18−x)=124 ⇒2x182−x2=124 ⇒48x=324−x2 ⇒x2+48x−324=0 ⇒x2+54x−6x−324=0 ⇒x(x+54)−6(x+54)=0 ⇒(x+54)(x−6)=0 ⇒x=−54 or x=6 But as x>0, so x can't be negative. Hence, x=6.

Solve: $\frac{1}{18 - x} - \frac{1}{18 + x} = \frac{1}{24}$ and $x > 0$ .