Prove that ax² + bx + c = 0 and cy² + by + a = 0 have reciprocal roots:

Answer 1

Given ax² + bx + c = 0, a =/= 0, and c =/= 0 we have x = [ -b +- √(b² - 4ac) ] / 2a,

and given cy² + by + a = 0, c =/= 0, and a =/= 0 we have y = [ -b +- √(b² - 4ca) ] / 2c,

so now we have the following:

y = { [ -b +- √(b² - 4ca) ][ -b -+ √(b² - 4ca) ] } / { 2c [ -b -+ √(b² - 4ca) ] }

y = { [-b]² - [√(b² - 4ca)]² } / { 2c [ -b +- √(b² - 4ac) ] }

y = { b² - (b² - 4ca) } / { 2c [ -b +- √(b² - 4ac) ] }

y = { b² - b² + 4ca } / { 2c [ -b +- √(b² - 4ac) ] }

y = { 4ca } / { 2c [ -b +- √(b² - 4ac) ] } = 2a / [ -b +- √(b² - 4ac) ] = 1/x.