Find the quadratic equation whose roots are half of the reciprocal of the roots of the equation $\displaystyle ax^{2}+bx+c=0$
- $\displaystyle 4ax^{2}+2bx+c=0$
- $\displaystyle 4cx^{2}+2bx+a=0$
- $\displaystyle 2cx^{2}+bx+a=0$
- $\displaystyle 2ax^{2}+bx+c=0$
If the roots of ax^2 + bx + c = 0 are r1 and r2, the new roots are 1/(2r1) and 1/(2r2). Let y = 1/(2x), so x = 1/(2y). Substituting this into the original equation gives a(1/(2y))^2 + b(1/(2y)) + c = 0, which simplifies to a/(4y^2) + b/(2y) + c = 0. Multiplying by 4y^2 yields a + 2by + 4cy^2 = 0.
Let the roots of the original equation ax^2 + bx + c = 0 be p and q. By Vieta's formulas, the sum of the roots is p + q = -b/a and the product of the roots is pq = c/a. The new roots are 1/(2p) and 1/(2q), so their sum is (1/2)(1/p + 1/q) = (p + q)/(2pq) = (-b/a) / (2c/a) = -b/(2c). The product of the new roots is 1/(4pq) = 1 / (4c/a) = a/(4c). The required quadratic equation is x^2 - (sum)x + (product) = 0, which becomes x^2 + (b/2c)x + a/(4c) = 0. Multiplying the entire equation by 4c yields the equivalent equation 4cx^2 + 2bx + a = 0.