## Formation of Quadratic Equations With Given Roots:

Let α and β be the roots of the equation ax^{2} + bx + c = 0, where a ≠ 0. Then, α + β = -b/a and αβ = c/a.

Now, the equation can be written as-

x^{2} +(b/a)x + c/a = 0 or, x^{2} – (-b/a)x + c/a = 0

⇒ x^{2} – (α + β)x + αβ = 0 ……….(i)

⇒ x^{2} – (sum of the roots). x + product of the roots = 0

Thus, the equation whose roots are α and β may be written as x^{2} – (α + β)x + αβ = 0.

Equation (i) can also be written as-

x^{2} – αx – βx + αβ = 0

⇒ x (x – α) – β (x – α) = 0

⇒ (x – α) (x – β) = 0

Thus, the equation with α and β as roots can also be written as (x – α) (x – β) = 0.

Example- If α, β are the roots of the equation x^{2} – 2x + 5 = 0 then form an equation whose roots are (i) 2α, 2β (ii) 1/α, 1/β (iii) α^{2}, β^{2}Solution- The given equation is,x ^{2} – 2x + 5 = 0Now, α and β are its roots. ∴ α + β = 2 and αβ = 5 (i) Now the roots of the required equation are 2α, 2β∴ Sum of roots = 2α + 2β = 2 (α + β) = 2(2) = 4 and Product of roots = (2α) (2β) = 4αβ = 4(5) = 20 The required equation is, x ^{2} – 4x + 20 = 0(ii) Now the roots of the required equation are 1/α, 1/β∴ Sum of roots = 1/α + 1/β = (β + α)/αβ = 2/5 and Product of roots = (1/α) (1/β) = 1/αβ = 1/5 The required equation is, x ^{2} – (2/5)x + 1/5 = 05x ^{2} – 2x + 1 = 0(iii) Now the roots of the required equation are α^{2}, β^{2}∴ Sum of roots = α ^{2}+ β^{2} = (α + β)^{2} – 2αβ = (2)^{2} – 2(5) = 4 – 10 = -6and Product of roots = α ^{2}x β^{2} = α^{2}β^{2} = (5)^{2} = 25The required equation is, x ^{2} + 6x + 25 = 0 |

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