Question

х2- ax/b + x2 - bx/a + x2 - Зах – 3bx/
a + b
= 0​

Answer 1

Question

$\large\bf \: \frac{ x {}^{2} - \frac{ax}{b} + x {}^{2} - \frac{bx}{a} + x {}^{2} - 3ax - 3bx}{a + b} = 0$

Answer

$\large\bf \: \frac{ x {}^{2} - \frac{ax}{b} + x {}^{2} - \frac{bx}{a} + x {}^{2} - 3ax - 3bx}{a + b} = 0$

$\mapsto\tt \: \frac{ \cancel{x {}^{2}} - \frac{ax}{b} + \cancel{x {}^{2}} - \frac{bx}{a} + x {}^{2} - 3ax - 3bx}{a + b} = 0$

$\mapsto \: \tt \: \frac{ - \frac{ax}{b} - \frac{bx}{a} + x {}^{2} - 3ax - 3bx}{a + b} = 0$

$\mapsto \: \tt \: \frac{ - \frac{a(ax) - b(bx)}{a \times b} + x {}^{2} - 3ax - 3bx}{a + b} = 0$

$\mapsto \: \tt \: \frac{ - \frac{a {}^{2}x - b {}^{2}x }{ab} + x {}^{2} - 3x (a-b) }{a + b} = 0$

$\mapsto \: \tt \: \frac{ - \frac{x(a {}^{2} - b {}^{2} )}{ab} + x {}^{2} - 3x (a-b) }{a + b}0 = 0$

$\mapsto \: \tt \: \frac{ - x(a {}^{2} - b {}^{2} ) + x {}^{2} - 3x (a-b) }{a {}^{2} + b {}^{2} } = 0$

$\mapsto \: \tt \: \frac{ - {x(\cancel{a {}^{2} - b {}^{2} )} } + x {}^{2} - 3x (a-b) } { \cancel{a {}^{2} + b {}^{2} }} = 0$

$\tt - x + x {}^{2} - 3x(a - b) = 0$

$\tt x {}^{2} - 4x(a - b) = 0$

Either,

x²-4x=0

x² = 4x

x = √4x

x = 2x

0r,

a-b=0

a=b