Let a- b- c be the positive real numbers such that bx2-a-c2-4b2x- a-c - 0 for all x belongs to R- then a- b- c are in

Explanation for the correct option:Given that,bx^2+√(((a+c)^2+4b^2))x+ a+c ≥ 00ex0ex⇒ bx^2+((a+c)^2+4b^2)^0.5x+ a+c ≥ 0 ....(i)Since ∀ x is a real number ...

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Byju's Answer

Standard XI

Mathematics

Properties of Inequalities

Let a, b, c b...

Question

Let $a, b, c$ be the positive real numbers such that $b x^{2} + \sqrt{({(a + c)}^{2} + 4 b^{2})} x + a + c \geq 0$ for all $x$ belongs to $R$ , then $a, b, c$ are in

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Solution

The correct option is A
AP

Explanation for the correct option:
Given that,
$b x^{2} + \sqrt{({(a + c)}^{2} + 4 b^{2})} x + a + c \geq 0 \Rightarrow b x^{2} + ({{(a + c)}^{2} + 4 b^{2})}^{0.5} x + a + c \geq 0 . . . . (i)$
Since $\forall x$ is a real number, therefore
Discriminant of $(i)$ $\leq 0$
$\Rightarrow {(a + c)}^{2} + 4 b^{2} - 4 b (a + c) \leq 0 \Rightarrow (a^{2} + 2 a c + c^{2}) + 4 b^{2} - (4 a b - 4 b c) \leq 0 \Rightarrow {(a + c - 2 b)}^{2} \leq 0 \Rightarrow a + c - 2 b \leq 0$ \
Therefore,
$2 b = a + c$
Thus, $a, b, c$ are in AP.
Hence, the correct option is (A)

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Let a,b,c be the positive real numbers such that bx2+((a+c)2+4b2)x+a+c≥0 for all x belongs to R, then a,b,c are in

Let $a, b, c$ be the positive real numbers such that $b x^{2} + \sqrt{({(a + c)}^{2} + 4 b^{2})} x + a + c \geq 0$ for all $x$ belongs to $R$ , then $a, b, c$ are in