If x1​ and x2​ are the roots of the equation 2x2+3x−9=0 then the equation…

Concept

For a quadratic ax2+bx+c=0 with roots α and β, Vieta's formulas give the sum α+β = −b/a and the product αβ = c/a. To build a new quadratic whose roots are expressions of α and β, first compute the new sum S and new product P, then write x2 − Sx + P = 0.

Reciprocal-root shortcut: replacing x with 1/x in ax2+bx+c=0 (i.e. reversing the order of the coefficients) yields the equation whose roots are 1/α and 1/β, namely cx2+bx+a=0.

Application

1. Here a=2, b=3, c=−9, so the original roots satisfy x1+x2 = −b/a = −3/2 and x1x2 = c/a = −9/2.

2. New sum: 1/x1 + 1/x2 = (x1+x2)/(x1x2) = (−3/2)/(−9/2) = 1/3.

3. New product: 1/(x1x2) = 1/(−9/2) = −2/9.

4. Form the equation: x2 − (1/3)x + (−2/9) = 0, i.e. x2 − (1/3)x − 2/9 = 0.

5. Multiply through by 9 to clear the denominators: 9x2 − 3x − 2 = 0.

Cross-check

Apply the reciprocal shortcut directly: substitute x → 1/x in 2x2+3x−9=0 to get 2/x2 + 3/x − 9 = 0; multiplying by x2 gives 2 + 3x − 9x2 = 0, and multiplying by −1 gives 9x2 − 3x − 2 = 0 — the same equation, confirming the result.