Question

Solve the following quadratic equations by factorization:
(1/x+4)-(1/x-7)=11/30,x≠4,7

Answer 1

The values of x in the given quadratic equation is 1 and 2 are the roots

Step-by-step explanation:

Given quadratic equation is $\frac{1}{x+4}-\frac{1}{x-7}=\frac{11}{30}$ and $x\neq 4,7$

To solve the given equation by Factorization method :

• $\frac{1}{x+4}-\frac{1}{x-7}=\frac{11}{30}$  
• $\frac{1(x-7)-1(x+4)}{(x+4)(x-7)}=\frac{11}{30}$

• $\frac{x-7-1(x)-1(4)}{x(x)+x(-7)+4(x)+4(-7)}=\frac{11}{30}$
• $\frac{x-7-x-4}{x^2-7x+4x-28}=\frac{11}{30}$

• $\frac{-11}{x^2-3x-28}=\frac{11}{30}$
• $\frac{30}{x^2-3x-28}=\frac{11}{-11}$

• $\frac{30}{x^2-3x-28}=-1$
• $30=-1(x^2-3x-28)$
• $30=-x^2+3x+28$

Rewritting we get

• $-x^2+3x+28=30$

Dividing by "-" on both sides

• $x^2-3x-28=-30$

• $x^2-3x-28+30=0$

• $x^2-3x+2=0$

• $(x-1)(x-2)=0$

• x-1=0 or x-2=0

• x=1 or x=2

Therefore the values of x is 1 and 2