If a quadratic equation in the general form, ax² + bx + c = 0 has roots α and β, then (x – α)(x – β) = 0. Or expanding this will give x² – (α + β)x + αβ = 0. Comparing the two equations will result in α + β = − b/a, αβ = c/a.

**Sum of Roots : α + β = − b/a**

**Product of Roots : αβ = c/a**

Example :

If α, β are the roots of the equation, 2x² – 10x -1 = 0, form the equations whose roots are (a) 2α, 2β (b) α + 2, β + 2.

Solution:

α + β = − (−10/2) = 5, and αβ = (−1/2)

(a) Sum of roots = 2α + 2β = 2(α + β) = 2(5) = 10

Product of roots = 2α.2β = 4αβ = 4(−1/2) = −2

∴ equation is x² – (10)x + (−2) = 0 or x² − 10x −2 = 0

(b) Sum of roots = α + 2 + β + 2 = α + β + 4 = 5 + 4 = 9

Product of roots = (α + 2)(β + 2) = αβ + 2α +2β + 4 = (−1/2) + 2(5) + 4 = 27/2.

∴ equation is x² – (9)x + (27/2) = 0 or 2x² − 9x + 27 = 0.

*Now, have fun and try to solve the following questions*.

1) If α and β are the roots of the equation, 3x² + 6x – 5 = 0, then find the equation whose roots are 1/α and 1/β.

2) The roots of the equation, x² + 4x – h = 0 are α and (α + k). Express h in terms of k.

*Here are the answers* :

1) α + β = – 6/3 = -2, αβ = – 5/3

Sum of roots = 1/α + 1/β = (β +α)/(αβ) = -2/(-5/3) = 6/5

Product of roots = (1/α)(1/β) = 1/(αβ) = 1/(-5/3) = −3/5

∴ Equation is x² – (6/5)x + (−3/5) = 0 ⇒ 5x² – 6x − 3 = 0.

2) Sum of roots = α + α + k = −4 ⇒ 2α = −k −4 ⇒ α = (−k −4)/2.

Product of roots = α(α + k) = −h.

Substitute α into the product equation :

〈(−k −4)/2〉 〈(−k −4)/2 + k)〉 = −h

〈(−k −4)/2〉 〈(−k −4 +2k)/2〉 = −h

〈(−k −4)/2〉 〈(k −4)/2〉 = −h

h = (k + 4)(k − 4)/4

h = (k² − 16)/4

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