Question

Find the value of 'p' for which px²+6x+2=0 has both real roots.
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Answer 1

Answer:

p = 4.5

Step-by-step explanation:

Dear this criteria is only possible when the eqn. have two equal and real roots ..

hope this helps dear

Answer 2

$\huge \fbox \pink{Answer:}$

$\textbf{Given:- } \text{Equation is } \: P {x}^{2} + 6x + 1 = 0 \: \text{with \: real \: roots}$

$\textbf{To Find:- } \: \text{Maximum Value of P}$

$\textbf{Compare With} \: a {x}^{2} + bx + c = 0$

$\textbf{Here,} \: a = p, \: b = 6, \: c = 1$

$\textbf{We Know for Real Roots } \: {b}^{2} - 4ac \: \geqslant 0$

$⇒(6 {)}^{2} - 4(p)(1) \geqslant 0$

$⇒36 - 4p \geqslant 0$

$⇒36 \geqslant 4p$

$⇒p \leqslant 9$

• Hence The Maximum Value of p = 9

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