Find the value of $p$ for which given quadratic has real roots.
$5 x^{2} + 8 x + p = 0$

p \geq \frac{5}{16}

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p \leq \frac{5}{16}

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p \leq \frac{- 5}{16}

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p \leq \frac{16}{5}

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Solution

The correct option is D $p \leq \frac{16}{5}$
$5 x^{2} + 8 x + p = 0$
$Δ \geq 0, b^{2} - 4 a c \geq 0$
$a = 5, b = 8, c = p$
${(8)}^{2} - 4 (5) c, (p) \geq 0$
$64 - 20 p \geq 0$
$p \leq \frac{64}{20}$
$p \leq \frac{16}{5}$
$p \geq \frac{5}{16}$
$p \leq \frac{5}{16}$
$p \leq \frac{- 5}{16}$
$p \leq \frac{16}{5}$

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Find the value of p for which given quadratic has real roots.5x2+8x+p=0

The correct option is D p≤1655x2+8x+p=0Δ≥0,b2−4ac≥0a=5,b=8,c=p(8)2−4(5)c,(p)≥064−20p≥0p≤6420p≤165p≥516p≤516p≤−516p≤165

Find the value of $p$ for which given quadratic has real roots.
$5 x^{2} + 8 x + p = 0$