English

Solve the Following Quadratic Equations by Factorization: - Mathematics

Advertisements
Advertisements

Question

Solve the following quadratic equations by factorization: \[\frac{2}{x + 1} + \frac{3}{2(x - 2)} = \frac{23}{5x}; x \neq 0, - 1, 2\]

Answer in Brief
Advertisements

Solution

\[\frac{2}{x + 1} + \frac{3}{2(x - 2)} = \frac{23}{5x}\]

\[ \Rightarrow \frac{4(x - 2) + 3(x + 1)}{2(x - 2)(x + 1)} = \frac{23}{5x}\]

\[ \Rightarrow \frac{4x - 8 + 3x + 3}{2( x^2 + x - 2x - 2)} = \frac{23}{5x}\]

\[ \Rightarrow \frac{7x - 5}{2 x^2 - 2x - 4} = \frac{23}{5x}\]

\[ \Rightarrow 5x\left( 7x - 5 \right) = 23\left( 2 x^2 - 2x - 4 \right)\]

\[ \Rightarrow 35 x^2 - 25x = 46 x^2 - 46x - 92\]

\[ \Rightarrow 46 x^2 - 35 x^2 - 46x + 25x - 92 = 0\]

\[ \Rightarrow 11 x^2 - 21x - 92 = 0\]

\[ \Rightarrow 11 x^2 - 44x + 23x - 92 = 0\]

\[ \Rightarrow 11x(x - 4) + 23(x - 4) = 0\]

\[ \Rightarrow (11x + 23)(x - 4) = 0\]

\[ \Rightarrow 11x + 23 = 0 \text { or } x - 4 = 0\]

\[ \Rightarrow x = - \frac{23}{11} \text { or } x = 4\]

Hence, the factors are 4 and \[- \frac{23}{11}\].

shaalaa.com
  Is there an error in this question or solution?
Chapter 4: Quadratic Equations - Exercise 4.3 [Page 20]

APPEARS IN

RD Sharma Mathematics [English] Class 10
Chapter 4 Quadratic Equations
Exercise 4.3 | Q 35 | Page 20

RELATED QUESTIONS

Find the roots of the following quadratic equation by factorisation:

x2 – 3x – 10 = 0


Solve the following quadratic equations by factorization:

a2b2x2 + b2x - a2x - 1 = 0


There are three consecutive integers such that the square of the first increased by the product of the first increased by the product of the others the two gives 154. What are the integers?


The product of Ramu's age (in years) five years ago and his age (in years) nice years later is 15. Determine Ramu's present age.


Solve  2x2 – 9x + 10 =0; when x ∈ Q


Solve the following quadratic equations by factorization: 

`(x + 3)^2 – 4(x + 3) – 5 = 0 `


Solve the following quadratic equations by factorization: 

`4(2x – 3)^2 – (2x – 3) – 14 = 0`


The sum of the squares of two consecutive positive integers is 365. Find the integers. 

 


Solve the following quadratic equation by factorization: \[\frac{a}{x - b} + \frac{b}{x - a} = 2\]


Find the values of k for which the roots are real and equal in each of the following equation:

\[kx\left( x - 2\sqrt{5} \right) + 10 = 0\]


Solve the following equation:  4x2 - 13x - 12 = 0


Solve the following equation: `"a"("x"^2 + 1) - x("a"^2 + 1) = 0`


Solve the following equation:  `x^2 + (a + 1/a)x + 1 = 0`


Solve the following quadratic equation by factorisation:
9x2 - 3x - 2 = 0


Solve the equation:
`6(x^2 + (1)/x^2) -25 (x - 1/x) + 12 = 0`.


Find the values of x if p + 1 =0 and x2 + px – 6 = 0


Complete the following activity to solve the given quadratic equation by factorization method.

Activity: x2 + 8x – 20 = 0

x2 + ( __ ) – 2x – 20 = 0

x (x + 10) – ( __ ) (x + 10) = 0

(x + 10) ( ____ ) = 0

x = ___ or x = 2


Solve the following quadratic equation by factorisation method:

x2 + x – 20 = 0


4x2 – 9 = 0 implies x is equal to ______.


For equation `1/x + 1/(x - 5) = 3/10`; one value of x is ______.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×