Question

can anyone sopve question no. 4 and 6

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Answer 1

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Answer 2

[4.]

The given quadratic equation is

(k - 2)x² - (k - 4)x - 2 = 0

If α and β the roots of the given equation, then

α + β = - { - (k - 4)/(k - 2)} = (k - 4)/(k - 2)

αβ = - 2/(k - 2)

Given that, difference of the roots = 3

⇒ α - β = 3

Now, (α - β)²

= (α + β)² - 4αβ

= {(k - 4)/(k - 2)}² - 4 {- 2/(k  -2)}

= {(k - 4)² + 8 (k - 2)}/(k - 2)²

= (k² - 8k + 16 + 8k - 16)/(k² - 4k + 4)

= k²/(k² - 4k + 4)

⇒ 3² (k² - 4k + 4) = k²

⇒ 9k² - 36k + 36 - k² = 0

⇒ 8k² - 36k + 36  = 0

⇒ 2k² - 9k + 9 = 0

⇒ (2k - 3) (k - 3) = 0

Either 2k - 3 = 0 or, k - 3 = 0

⇒ k = 3/2 , k = 3

[6.]

The given equations are

ax² + bx + c = 0 .....(i)

dx² + ex + f = 0 .....(ii)

Since α and β are roots of (i), using quadratic formula, we can find the roots as

α = { - b + √(b² - 4ac) }/(2a)

β = { - b - √(b² - 4ac) }/(2a)

Since (α + 2) and (β + 2) are roots of (ii), using quadratic formula, we can find the roots as

α + 2 = { - e + √(e² - 4df) }/(2d)

β + 2 = { - e - √(e² - 4df) }/(2d)

Putting the values of α and β from the previous solved ones, we get

{ - b + √(b² - 4ac) }/(2a) + 2

= { - e + √(e² - 4df) }/(2d)

⇒ (- b)/(2a) + {√(b² - 4ac)}/(2a) + 2

= (- e)/(2d) + {√(e² - 4df)}/(2d) .....(iii)

{ - b - √(b² - 4ac) }/(2a) + 2

= { - e - √(e² - 4df) }/(2d)

⇒ (- b)/(2a) - {√(b² - 4ac)}/(2a) + 2

= (- e)/(2d) - {√(e² - 4df)}/(2d) .....(iv)

On subtraction, from (iii) and (iv), we get

2 {√(b² - 4ac)}/(2a) = 2 {√(e² - 4df)}/(2d)

⇒ √{(b² - 4ac)/(e² - 4df)} = a/d

Squaring both sides, we get

(b² - 4ac)/(e² - 4df) = a²/d²