## Roots and Coefficients of a Quadratic Equations:

The roots of the quadratic equation ax^{2} + bx + c = 0 (a ≠ 0) are given by,

Example 1- If one of the roots of the equation ax^{2} + bx + c = 0 is three times the other, then prove that 3b^{2} = 16 ac.Solution- The given equation is,ax ^{2} + bx + c = 0Let α, 3α be its roots. Now, α + 3α = -b/a ⇒ 4α = -b/a ⇒ α = -b/4a …………(i) Also, α . 3α = c/a ⇒ 3α ^{2} = c/a⇒ α ^{2} = c/3a …………(ii)Substitute the value of α from (i) in (ii)- (-b/4a) ^{2} = c/3a⇒ b ^{2}/16a^{2} = c/3a⇒ 3b ^{2} = 16ac |

Example 2- If αβ are the roots of the equation x^{2} – 6x + P = 0 such that 3α + 2β = 20, then find the value of P.Solution- The given equation is, x ^{2} – 6x + P = 0Now, α, and β are their roots. α + β = 6 ……….(i) and αβ = P ……….(ii) Also 3α + 2β = 20 ……….(iii) Solving (i) and (iii)- α + β = 6 3α + 2β = 20 ⇒ α/(20 – 12) = β/(18 – 20) = -1/(2 – 3) ⇒ α/8 = β/(-2) = 1/1 ⇒ α = 8, β = -2 Now from (ii)- αβ = P ⇒ (8) (-2) = P ⇒ P = -16 |

Example 3- If the difference of the roots of equation x^{2} + ax + b = 0 be unity then prove that a^{2} + 4b^{2} = (1 + 2b)^{2}.Solution- The given equation is, x ^{2} + ax + b = 0Let its roots be α and β such that α – β = 1 ⇒ √[(α + β) ^{2} – (4αβ)] = 1⇒ √[(-a) ^{2} – 4b] = 1Squaring both sides, we get- ⇒ a ^{2} – 4b = 1⇒ a ^{2} = 1 + 4b⇒ a ^{2} + 4b^{2} = 1 + 4b + 4b^{2}⇒ a ^{2} + 4b^{2} = (1 + 2b)^{2} |

Example 4- If k be the ratio of the roots of equation x^{2} – ax + b = 0 then prove that (k^{2} + 1)/k = (a^{2} – 2b)/b.Solution- The given equation is,x ^{2} – ax + b = 0Let αk and α be the roots of the equation. Sum of root = -b/a ⇒ αk + α = a ⇒ α (k + 1) = a ⇒ α = a/(k + 1) ……….(i) Product of root = c/a ⇒ αk . α = b ⇒ α ^{2}k = b ……….(ii)Substitute (i) in (ii), we get- [a/(k + 1)] ^{2} k = b⇒ [a ^{2}/(k^{2} + 1 + 2k)] k = b⇒ a ^{2}k = b (k^{2} + 1 + 2k)⇒ b (k ^{2} + 1) = a^{2}k – 2bk⇒ b (k ^{2} + 1) = k (a^{2} – 2b)⇒ (k ^{2} + 1)/k = (a^{2} – 2b)/b |