Question

Find the values of p for which the quadratic equation px² - 6x + p = 0 has real and distinct roots.
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Answer 1

The given equation is px² - 6x + p = 0.

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• Comparing with ax² + bx + c = 0

We get,

➣ a = p

➣ b = -6

➣ c = p

∴ Discriminant = b² - 4ac

➺ (-6)² - 4.p.p

➺ 36 - 4p².

• For real and different roots, discriminant > 0

➝ 36 - 4p² > 0

➝ 9 - p² > 0

➝ p² - 9 < 0

➝ (p + 3) (p - 3) < 0

➝ -3 < p < 3

❑ Therefore,

• Hence, the given equation has real and different roots if -3 < p < 3.

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✯ Note :

✠ If α, β are real numbers such that α < β, then -

• (x - α) (x - β) < 0 if and only if α < x < β.

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Answer 2

px² - 6x + p = 0

Comparing :-

a=p

b=-6

c=p

Discriminant :- b²-4ac

(6)²-4(p)(p)

36-4p²>0

9-p²>0

p²-9<0

(p+3)(p+3)<0

Hence the roots of the given equation are real and different -3<p<3