Title Ep-4(B) Quadratic Equations Solutions.pdf ✅

Type-1 : Manipulations Based 1. Video 2. We have, a + b + c = 1, ab + bc + ca = 2 and abc = 3 Now, a 2 + b 2 + c 2 = (a + b + c) 2 − 2Σab = −3 ⇒ (ab + bc + ca) 2 = Σ(ab) 2 + 2abcΣa ⇒ Σ(ab) 2 = −2 So, a 4 + b 4 + c 4 = a 2 + b 2 + c 2 2 − 2Σ(ab) 2 = 9 − 2(−2) = 13 3. Type(14) 137 (4) Given p + q = 3 and p 4 + q 4 = 369 1 p + 1 q −2 = pq p + q 2 = (qp) 2 9 Now, (p + q) 2 = 9 ⇒ p 2 + q 2 = 9 − 2pq p 4 + q 4 = (p 2 + q 2 ) 2 − 2p 2 q 2 ⇒ 369 = (9 − 2pq) 2 − 2(pq) 2 369 = 81 + 4p 2 q 2 − 36pq − 2p 2 q 2 ⇒ 288 = 2p 2 q 2 − 36pq ⇒ 144 = p 2 q 2 − 18pq (pq) 2 − 2 × 9 × pq + 92 = 144 + 92 (pq − 9)2 = 225 ⇒ pq − 9 = ±15 ⇒ pq = ±15 + 9 pq = 24, −6 ∵ p 2 + q 2 ≥ 0 So 24 is rejected (qp) 2 9 = 16×16 9 = 4 4. Consider (p 2 + q 2 ) 2 − 2p 2 q 2 = 272 (p + q) 2 − 2pq2 − 2p 2 q 2 = 272 16 − 16pq + 2p 2 q 2 = 272 (pq) 2 − 8pq − 128 = 0 (pq) 2 − 8pq − 128 = 0 pq = 8 ± 24 2 = 16, −8 ∴ pq = 16
∴ Required equation : x 2 − (2)x + 16 = 0 Type-2 : Applying α ± β; αβ formulae directly 5. Type(11) 203 Let α and β are roots of the equation x 2+ax+b = 0 and γ and δ are roots of the equation x 2 +bx+a = 0 respectively. α+β = −a, αβ = b and γ +δ = −b, γδ = a Given |α − β| = |γ − δ| ⇒ (α − β) 2 = (γ − δ) 2 ⇒ (α + β) 2 − 4αβ = (γ + δ) 2 − 4γδ ⇒ a 2 − 4b = b 2 − 4a ⇒ (a 2 − b 2 ) + 4(a − b) = 0 ⇒ a + b + 4 = 0 (∵ a ̸= b) 6. Type(11) 187 (c) Given quadratic eqn. is x 2 + px + 3p 4 = 0 So, α + β = −p, αβ = 3p 4 Now, given |α − β| = √ 10 ⇒ α − β = ± √ 10 ⇒ (α − β) 2 = 10 ⇒ α 2 + β 2 − 2αβ = 10 ⇒ (α + β) 2 − 4αβ = 10 ⇒ p 2 − 4 × 3p 4 = 10 ⇒ p 2 − 3p − 10 = 0 ⇒ p = −2, 5 ⇒ p ∈ {−2, 5} 7. Type(11) 200 (d) Given that 4 is a root of x 2 + px + 12 = 0 ⇒ 16 + 4p + 12 = 0 ⇒ p = −7 Now, the equation x 2 + px + q = 0 has equal roots. ∴ D = 0 ⇒ p 2 − 4q = 0 ⇒ q = p 2 4 = 49 4 8. Type(11) 201 (c) Let the second root be α. Then α + (1 − p) = −p ⇒ α = −1 Also α.(1 − p) = 1 − p ⇒ (α − 1)(1 − p) = 0 ⇒ p = 1[∵ α = −1] 9. Type(11) 120 (45) Given, x 2 + 60 1 4 x + a = 0 with roots α&β. Given α 4 + β 4 = −30 α + β = −60 1 4&αβ = a α 4 + β 4 = (α 2 + β 2 ) 2 − 2α 2β 2 Since, (α 2 + β 2 ) 2 = ((α + β) 2 − 2αβ) 2 Page 2
⇒ {(α + β) 2 − 2αβ} 2 − 2a 2 = −30 ⇒ n 60 1 2 − 2a o2 − 2a 2 = −30 ⇒ 60 + 4a 2 − 4a × 60 1 2 − 2a 2 = −30 ⇒ 2a 2 − 4 · 60 1 2 a + 90 = 0 Product = 90 2 = 45 10. Type(11) 126 (b) For x 2 + ax + b = 0 (Sum of roots) = a = −1 α2 − 1 β2 − 2 = a (Product of roots) = b = 1 α2 + 1 β2 + 1 + 1 α2β2 a + b = 1 (αβ) 2 − 1 = 1 6 − 1 = − 5 6 [∵ αβ = √ 6] 11. Type(11) 138 (b) Given α, β roots of equation 3x 2 + λx − 1 = 0 So, α + β = −λ 3 , αβ = −1 3 According to given relation of roots, 1 α2 + 1 β2 = (α+β) 2−2αβ α2β2 = 15 λ 2 = 9 ⇒ λ = ±3 Now, α 2 + β 2 = (α + β) 2 − 2αβ α 2 + β 2 = ±3 3 2 + 2 3 = 5 3 Take, 6 (α 3 + β 3 ) 2 = 6 ((α + β) (α 2 + β 2 − αβ)) = 6 − (±3) 3 5 3 + 1 3 2 = 6(1 × 2)2 = 6 × 4 = 24. 12. Type(11) 150 (a) ∵ α + β = 64, αβ = 256 α 3/8 β5/8 + β 3/8 α5/8 = α+β (αβ) 5/8 = 64 (28) 5/8 = 64 32 = 2 13. Type(11) 153 (d) Let α and β be the roots of the quadratic equation 7x 2 − 3x − 2 = 0 ∴ α + β = 3 7 , αβ = −2 7 Now, α 1−α2 + β 1−β2 = α−αβ(α+β)+β 1−(α2+β2)+(αβ) 2 = (α+β)−αβ(α+β) 1−(α+β) 2+2αβ+(αβ) 2 = 3 7 + 2 7 × 3 7 1− 9 49 +2× −2 7 + 4 49 = 27 16 14. Type(11) 155 (d) α · β = 2 and α + β = −p also 1 α + 1 β = −q ⇒ p = 2q Now α − 1 α β − 1 β α + 1 β β + 1 α = h αβ + 1 αβ − α β − β α i hαβ ̇ + 1 αβ + 2i Page 3
15. Type(11) 163 (d) Since 2 − √ 3 is a root of the quadratic equation x 2 + px + q = 0 ∴ 2 + √ 3 is the other root Sum of roots = −p = (2 − √ 3) + (2 + √ 3) = 4 ⇒ p = −4 Product of roots = q = (2 − √ 3)(2 + √ 3) = 1 ⇒ q = 1 ⇒ p 2 − 4q − 12 = 16 − 4 − 12 = 0 16. Type(11) 166 (b) Let roots of the quadratic equation are α, β. Given, λ = α β and λ + 1 λ = 1 ⇒ α β + β α = 1 (α+β) 2−2αβ αβ = 1 The quadratic equation is, 3m2x 2 + m(m − 4)x + 2 = 0 ∴ α + β = m(4−m) 3m2 = 4−m 3m and αβ = 2 3m2 Put these values in eq (i), 4−m 3m 2 − 4 3m2 2 3m2 = 3 ⇒ (m − 4)2 = 18 ⇒ m = 4 ± √ 18 Therefore, least value is 4 − √ 18 = 4 − 3 √ 2 17. Type(11) 167 (d) Let α and β be the roots of the equation, 81x 2 + kx + 256 = 0 Given α = β 3 ∵ Product of the roots = 256 81 ∴ (α)(β) = 256 81 ⇒ β 4 = 4 3 4 ⇒ β = 4 3 ⇒ α = 64 27 ∵ Sum of the roots = − k 81 ∴ α + β = − k 81 ⇒ 4 3 + 64 27 = − k 81 ⇒ k = −300 18. Type(11) 172 (b) 1 x+p + 1 x+q = 1 r ⇒ x+p+x+q (x+p)(x+q) = 1 r (2x + p + q)r = x 2 + px + qx + pq x 2 + (p + q − 2r)x + pq − pr − qr = 0 Let α and β be the roots. ∴ α + β = −(p + q − 2r) &αβ = pq − pr − qr ∵ α = −β (given); ∴ in eq. (i), we get ⇒ α + β = −(p + q − 2r) = 0 ⇒ p + q = 2r Now, α 2 + β 2 = (α + β) 2 − 2αβ From (ii) and (iii) = −2(pq − pr − qr) = −2pq + 2r(p + q) = p 2 + q 2 Page 4