Solve each of the following quadratic equations-12 a-b-2 x-12 a-1 b-12 x -CBSE 2013-

12a+b+2x=12a+1b+12x #8658;12a+b+2x 12x=12a+1b #8658;2x 2a b 2x2x2a+b+2x=2a+b2ab #8658; 2a+b4x2+4ax+2bx=2a+b2ab #8658;4x2+4ax+2bx= 2ab #8658;4x2+4ax+2 ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard X

Mathematics

Solving a Quadratic Equation by Factorization Method

Solve each of...

Question

Solve each of the following quadratic equations:

$\frac{1}{2 a + b + 2 x} = \frac{1}{2 a} + \frac{1}{b} + \frac{1}{2 x}$ [CBSE 2013]

Open in App

Solution

$\frac{1}{2 a + b + 2 x} = \frac{1}{2 a} + \frac{1}{b} + \frac{1}{2 x} \Rightarrow \frac{1}{2 a + b + 2 x} - \frac{1}{2 x} = \frac{1}{2 a} + \frac{1}{b} \Rightarrow \frac{2 x - 2 a - b - 2 x}{2 x (2 a + b + 2 x)} = \frac{2 a + b}{2 a b} \Rightarrow \frac{- (2 a + b)}{4 x^{2} + 4 a x + 2 b x} = \frac{2 a + b}{2 a b}$
$\Rightarrow 4 x^{2} + 4 a x + 2 b x = - 2 a b \Rightarrow 4 x^{2} + 4 a x + 2 b x + 2 a b = 0 \Rightarrow 4 x (x + a) + 2 b (x + a) = 0 \Rightarrow (x + a) (4 x + 2 b) = 0$
$\Rightarrow x + a = 0 or 4 x + 2 b = 0 \Rightarrow x = - a or x = - \frac{b}{2}$
Hence, $- a$ and $- \frac{b}{2}$ are the roots of the given equation.

Suggest Corrections

Similar questions

Q. Solve the following quadratic equations by factorization:

\frac{1}{2 a + b + 2 x} = \frac{1}{2 a} + \frac{1}{b} + \frac{1}{2 x}

Q. Solve each of the following quadratic equations:

\frac{1}{x + 1} + \frac{2}{x + 2} = \frac{5}{x + 4}, x \neq - 1, - 2, - 4

[CBSE 2013C]

Q. Solve each of the following quadratic equations:

3 (\frac{3 x - 1}{2 x + 3}) - 2 (\frac{2 x + 3}{3 x - 1}) = 5, x \neq \frac{1}{3}, - \frac{3}{2}

[CBSE 2014]

Q. Solve the following equations:

\frac{x + \sqrt{12 a - x}}{x - \sqrt{12 a - x}} = \frac{\sqrt{a + 1}}{\sqrt{a - 1}}

Q. Solve each of the following Cryptarithm:

\begin{matrix} 1 & 2 & A \\ + & 6 & A & B \end{matrix} \bar{\begin{matrix} A & 0 & 9 \end{matrix}}

Join BYJU'S Learning Program

Solve each of the following quadratic equations: 12a+b+2x=12a+1b+12x [CBSE 2013]

12a+b+2x=12a+1b+12x⇒12a+b+2x-12x=12a+1b⇒2x-2a-b-2x2x2a+b+2x=2a+b2ab⇒-2a+b4x2+4ax+2bx=2a+b2ab ⇒4x2+4ax+2bx=-2ab⇒4x2+4ax+2bx+2ab=0⇒4xx+a+2bx+a=0⇒x+a4x+2b=0 ⇒x+a=0 or 4x+2b=0⇒x=-a or x=-b2 Hence, -a and -b2 are the roots of the given equation.

Solve each of the following quadratic equations:

$\frac{1}{2 a + b + 2 x} = \frac{1}{2 a} + \frac{1}{b} + \frac{1}{2 x}$ [CBSE 2013]