Question

Solve the following quadratic equations by factorization:
(x-1/2x+1)+(2x+1/x-1)=5/2,x≠1/2,1

Answer 1

Answer:

x=-1

Step-by-step explanation:

• (x-1/2x+1) + (2x+1/x-1) = 5/2
• [(x-1)²+(2x+1)²] / [(2x+1)(x-1)] = 5/2
• [(x²-2x+1)+(4x²+4x+1)] / [2x²-x-1] = 5/2
• Adding the common terms we get,
• (5x²+2x+2) / (2x²x-1) = 5/2
• Cross multiplying,
• 10x²+4x+4 = 10x²-5x-5
• Taking all the terms on one side,
• 9x+9=0
• 9(x+1)=0
• x+1=0
• x=-1.

Answer 2

x = -1

Step-by-step explanation:

$\frac{(x-1)(x-1)+(2x+1)(2x+1)}{(2x+1)(x-1)} =\frac{5}{2} \\\frac{x^2-x-x=+14x^2+2x+2x+1}{2x^2-2x+x-1} =\frac{5}{2}\\\frac{x^2-2x+1+4x^2+4x+1}{2x^2-x-1} =\frac{5}{2}\\\frac{5x^2+2x+2}{2x^2-x-1}=\frac{5}{2}$

cross multiply

$2(5x^2+2x+2)=5(2x^2-x-1)\\10x^2+4x+4=10x^2-5x-5\\4x+5x+4+5=0\\9x+9=0\\9x=-9\\x=\frac{-9}{9} \\x=-1$