For all $`x^{'},$ $x^{2} + 2ax + (10 - 3a) 0$, then the interval in which $`a'$ lies, is

Question

For all $`x^{'},$ $x^{2} + 2ax + (10 - 3a) > 0$, then the interval in which $`a'$ lies, is

Dr. Abubacker · Accepted Answer

(b)
As we know, $ax^{2} + bx + c > 0\ for\ all\ x \in R,\ iff\ a > 0\ and\ D < 0$
$\therefore x^{2} + 2ax + (10 - 3a) > 0,\ \forall x \in R$
$\Rightarrow \ D < 0$
$\Rightarrow 4a^{2} - 4(10 - 3a) < 0$
$\Rightarrow 4\left( a^{2} + 3a - 10 \right) < 0$
$\Rightarrow (a + 5)(a - 2) < 0$
Using number line rule
[image:5_image7.emf]
$a \in ( - 5,\ 2)$

Q. For all $`x^{'},$ $x^{2} + 2ax + (10 - 3a) > 0$, then the interval in which $`a'$ lies, is

Explanation

Reviewed by Dr. Abubacker

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