If the equation $2 x^{2} - 6 x + p = 0$ has real and different roots, then the values of $p$ are given by

p < \frac{9}{2}

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p \leq \frac{9}{2}

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p > \frac{9}{2}

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p \geq \frac{9}{2}

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Solution

The correct option is A $p < \frac{9}{2}$
For having distinct real roots,
Discriminant, $D = b^{2} - 4 a c > 0 \Rightarrow (- 6)^{2} > 4 \cdot 2 \cdot p \Rightarrow 36 > 8 p$
$\Rightarrow 8 p < 36 \Rightarrow p < \frac{9}{2}$

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then the value of r is

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If the equation 2x2−6x+p=0 has real and different roots, then the values of p are given by

The correct option is A p<92For having distinct real roots, Discriminant, D=b2−4ac>0⇒(−6)2>4⋅2⋅p⇒36>8p⇒8p<36⇒p<92

If the equation $2 x^{2} - 6 x + p = 0$ has real and different roots, then the values of $p$ are given by

The correct option is A $p < \frac{9}{2}$
For having distinct real roots,
Discriminant, $D = b^{2} - 4 a c > 0 \Rightarrow (- 6)^{2} > 4 \cdot 2 \cdot p \Rightarrow 36 > 8 p$
$\Rightarrow 8 p < 36 \Rightarrow p < \frac{9}{2}$