Question

The roots of the equation x2+ (2p–1)x + p= 0 are real if.(a) p ≥1 (b) p ≤4 (c) p ≥1/4 (d) p ≤¼

Answer 1

Answer :

Given :

The quadratic equation is :

• x² + (2p - 1)x + p
• The roots of the given equation are real

To Find :

• The value of p for which the roots are real

Concept to be Used :

• The concept of discriminant is required to solve these kind problems.
• Discriminant of a quadratic equation is given by :
• $\sf b^{2} - 4ac$
• If Discriminant , b² - 4ac ≥0 then the roots of the equation are real
• If Discriminant, b² - 4ac=0 then the roots of the equation are real and equal
• If Discriminant , b² - 4ac<0 then the roots of the equation are immaginary ( or no real root exists)

Solution :

Given equation :

$\sf x^{2} + (2p - 1)x + p = 0$

Here : a = 1 , b = (2p - 1) and c = p

Also the roots of the equation are real so ,

$\sf \implies (2p - 1)^2 - 4\times 1 \times p \geqslant 0$

$\sf\implies 4p^{2} + 1 - 4p - 4p \geqslant 0 \\\\ \sf \implies 4p^{2} - 8p \geqslant -1 \\\\ \sf \implies 4p (p - 4) \geqslant -1 \\\\ \sf Thus \: \: we \: \: get \\ \sf 4p \geqslant -1 \: \: and \: \: p - 4 \geqslant -1 \\\\ \sf \implies p \geqslant \dfrac{-1}{4} \: and \: \implies p \geqslant 3$

Answer 2

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$\huge\tt{PROBLEM:}$

• The roots of the equation x2+ (2p–1)x + p= 0 are real if.(a) p ≥1 (b) p ≤4 (c) p ≥1/4 (d) p ≤¼

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$\huge\tt{SOLUTION:}$

x² + (2p-1)x + p = 0

Here, a = 1 , b = (2p-1) ,c = p

and roots of the equation are real,

↪(2p-1)² - 4 × 1 × p ≥ 0

↪4p² + 1 - 4p - 4p ≥ 0

↪4p² - 8p ≥ -1

↪4p(p-4) ≥ -1

↪p ≥ -¼

&

↪p ≥ 3

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