Quadratic Equations

6 )

The solution set of the equation (x + 1) (x + 2) (x + 3) (x + 4) = 120 is_______

[â€“ 6, 1 ({â€“ 5 Â± âˆš(39i)} / 2)]

[6, â€“ 1 {(â€“ 5 Â± âˆš39i) / 2}]

[â€“ 6, â€“ 1 {(â€“ 5 Â± âˆš39i) / 2}]

[6, 1 {(â€“ 5 Â± âˆš39i) / 2}]

View Solution

Answer :

(x + 1) (x + 2) (x + 3) (x + 4) = 120

∴ (x² + 5x + 4) (x² + 5x + 6) = 120

Let x² + 5x = m

∴ (m + 4) (m + 6) = 120

∴ m² + 10m + 24 = 120

∴ m² + 10m – 96 = 0

∴ m² + 16m – 6m – 96 = 0

∴ m(m + 16) – 6(m + 16) = 0

m = 6 OR m = – 16

∴ x² + 5x = 6 OR x² + 5x = – 16

∴ x² + 5x – 6 = 0 OR x² + 5x + 16 = 0

∴ x = – 6, x = 1 (1)

For x² + 5x + 16 = 0

x = [{– 5 ± √(25 – 64)} / 2]

x = [{– 5 ± √(39i²)} / 2]

∴ x = [{– 5 ± √(39i)} / 2] (2)

from (1) & (2)

x = [– 6, 1 ({– 5 ± √(39i)} / 2})]

7 )

The solution set of the Equation x⁴ – 5x³ – 4x² – 5x + 1 = 0 is__________

[3 Â± 2 âˆš2, {(â€“ 1 Â± âˆš3i) / 2}]

[{(3 Â± âˆš2) / 2}, â€“ 1 Â± âˆš3i}]

[â€“ 3 Â± 2 âˆš2, {(1 Â± âˆš3i) / 2}]

[{(â€“ 3 Â± 2 âˆš2) / 2}, 1 Â± âˆš3i]

View Solution

Answer :

x⁴ – 5x³ – 4x² – 5x + 1 = 0

∴ x² – 5x – 4 – (5/x) + (1 / x²) = 0

∴ x² + (1 / x²) – 5 {x + (1/x)} – 4 = 0 (1)

Let x + (1/x) = m

∴ x² + (1 / x²) = m² – 2

∴ from (1),

m² – 2 – 5m – 4 = 0

∴ m² – 5m – 6 = 0

∴ m = 6, m = – 1

when m = 6 ⇒ x + (1/x) = 6

∴ x² – 6x + 1 = 0

∴ x = 3 ± 2√2
when m = – 1, ⇒ x + (1/x) = – 1

x² + x + 1 = 0

∴ x = [(– 1 ± √3i) / 2]

∴ Solution set is [3 ± 2 √2, {(– 1 ± √3i) / 2}]

8 )

The solution set of equation [(x – a) / (x – b)] + [(x – b) / (x – a)] = (a/b) + (b/a), (a ≠ b) is _________

{a â€“ b, 0}

{(a/b), 0}

{a + b, 0}

{ab, 0}

View Solution

Answer :

[(x – a) / (x – b)] + [(x – b) / (x – a)] = (a/b) + (b/a) (a ≠ b)

Let [(x – a) / (x – b)] = m

∴ m + (1/m) = [(a² + b²) / ab]

∴ abm² – (a² + b²)m + ab = 0

∴ (bm – a) (am – b) = 0

∴ m = (a/b) and m = (b/a)

if m = (a/b) ⇒ [(x – a) / (x – b)] = (a/b) ⇒ x = 0

if m = (b/a) ⇒ [(x – a) / (x – b)] = (b/a) ⇒ x = a + b

∴ Solution set {a + b, 0}

9 )

If α & β are roots of quadratic equation x² + 13x + 8 = 0 then the value of α⁴ + β⁴ =________

23281

23218

23128

23182

View Solution

Answer :

x² + 13x + 8 = 0

Let α & β be the roots

α + β = – 136

αβ = 8

(α + β)² – 2 αβ = α² + β²

∴ (– 13)² – 2 × 8 = α² + β²

∴ α² + β² = 169 – 16

∴ α² + β² = 153

α⁴ + β⁴ = (α² + β²)² – 2 α² β²

= (153)² – 2(8)²

= 23281

10 )

The quadratic equation with rational, coefficient the sum of the squares of whose roots is 40 and the sum of the cubes of whose root is 208 is______

x2 + 4x + 12 = 0

x2 â€“ 4x â€“ 12 = 0

x2 â€“ 4x + 12 = 0

x2 + 4x â€“ 12 = 0

View Solution

Answer :

Let α and β be the roots.

α² + β² = 40

α² + β³ = 208

α + β = m and αβ = n

α² + β² = (α + β)² – 2 αβ

∴ 40 = m² – 2n

∴ [(m² – 40) / 2] = n

∴ α³ + β³ = 208

⇒ (α + β) (α² + β² – αβ) = 208

∴ m (40 – n) = 208

∴ m(40 – {(m² – 40) / 2}) = 208

∴ m³ – 120m + 416 = 0

∴ (m – 4) (m² + 4m – 104) = 0

∴ m – 4 = 0 and m² + 4m – 104 = 0

∴ m = 4 hence n = {(m² – 40) / 2} = – 12

∴ α + β = 4

αβ = – 12

Hence quadratic eqⁿ: x² – 4x – 12 = 0

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