The number of integral values of m for which | vedclass

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(C) For a quadratic expression $ax^2 + bx + c$ to be always positive for all $x \in R$,two conditions must be satisfied: $a > 0$ and $D < 0$.$1$. Cond

Vedclass · Accepted Answer

$7$

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(C) For a quadratic expression $ax^2 + bx + c$ to be always positive for all $x \in R$,two conditions must be satisfied: $a > 0$ and $D $1$. Condition $a > 0$:$(1 + 2m) > 0 \Rightarrow 2m > -1 \Rightarrow m > -\frac{1}{2}$.$2$. Condition $D $D = [-2(1 + 3m)]^2 - 4(1 + 2m)(4(1 + m)) $4(1 + 3m)^2 - 16(1 + 2m)(1 + m) $(1 + 6m + 9m^2) - 4(1 + 3m + 2m^2) $1 + 6m + 9m^2 - 4 - 12m - 8m^2 $m^2 - 6m - 3 Solving $m^2 - 6m - 3 = 0$ using the quadratic formula: $m = \frac{6 \pm \sqrt{36 - 4(1)(-3)}}{2} = \frac{6 \pm \sqrt{48}}{2} = 3 \pm \sqrt{12} = 3 \pm 2\sqrt{3}$.Since $\sqrt{12} \approx 3.46$,the roots are $3 - 3.46 = -0.46$ and $3 + 3.46 = 6.46$.So,$3 - 2\sqrt{3} Combining with $m > -0.5$,the interval is $(-0.46, 6.46)$.The integral values of $m$ are ${0, 1, 2, 3, 4, 5, 6}$.There are $7$ such values.

The number of integral values of $m$ for which the quadratic expression $(1 + 2m)x^2 - 2(1 + 3m)x + 4(1 + m)$ is always positive for all $x \in R$ is:

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