If-the-roots-of-2x-2-7x-5-0-are-the-reciprocal-roots-of-ax-2-bx-c-0-then-a-c-

Question Number 8662 by tttttyyyyuuuuuu77 last updated on 20/Oct/16

If the roots of 2 x^{2} + 7 x + 5 = 0 are the

reciprocal roots of {a x}^{2} + b x + c = 0,

then a - c =______.

Answered by sandy_suhendra last updated on 20/Oct/16

{2 x}^{2} + 7 x + 5 = 0 \Rightarrow the roots are α and β

so α + β = \frac{- 7}{2} and α β = \frac{5}{2}

{ax}^{2} + bx + c = 0 \Rightarrow the roots are x_{1} and x_{2}

x_{1} = \frac{1}{α} and x_{2} = \frac{1}{β}

x_{1} + x_{2} = \frac{1}{α} + \frac{1}{β} = \frac{α + β}{α β} = \frac{- 7 / 2}{5 / 2} = \frac{- 7}{5}

x_{1} x_{2} = \frac{1}{α β} = \frac{2}{5}

{ax}^{2} + bx + c = x^{2} - (x_{1} + x_{2}) x + x_{1} x_{2} = 0

x^{2} + \frac{7}{5} x + \frac{2}{5} = 0 \Rightarrow {5 x}^{2} + 7 x + 2 = 0

so a = 5, b = 7, c = 2

a - c = 5 - 2 = 3

Commented by sandy_suhendra last updated on 20/Oct/16

more simple

{2 x}^{2} + 7 x + 5 = 0 \Rightarrow the roots are α and β

{ax}^{2} + bx + c = 0 \Rightarrow the roots are x_{1} and x_{2}

x_{1} = \frac{1}{α} or α = \frac{1}{x} substitute to {2 x}^{2} + 7 x + 5 = 0

we have \Rightarrow 2 (\frac{1}{x^{2}}) + 7 (\frac{1}{x}) + 5 = 0

\frac{2}{x^{2}} + \frac{7}{x} + 5 = 0 \Rightarrow multiplied by x^{2}

2 + 7 x + {5 x}^{2} = 0

a = 5, b = 7 and c = 2