Consider the curves given by the following quadratic functions.
$⨍_{1} {(x)=5x}^{2} {+2x+1; ⨍}_{2} {(x)=5x}^{2} +6x+1 ⨍_{3} {(x)=x}^{2} {-7x+6; ⨍}_{4} {(x)=64x}^{4} +48x+9$
If $A_{1}, A_{2}, A_{3},$ and $A_{4}$ denote the lengths of the intercepts on the X-axis made by the above curves respectively, then which of the following is true ?

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$A_{3} > A_{2} > A_{4}$

$A_{4} > A_{2} > A_{3}$

$A_{2} > A_{4} > A_{3}$

$A_{1} > A_{2} > A_{3} > A_{4} > 0$

answer is C.

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Detailed Solution

$\begin{array}{l} T h e l e n g t h o f X - int e r c e p t o f a x^{2} + b x + c i s \sqrt{\frac{D}{| a |}} \\ f_{1} (x) = 5 x^{2} + 2 x + 1 \Rightarrow A_{1} = \frac{\sqrt{4 - 4 (5)}}{| 5 |} = \frac{\sqrt{- 16}}{5} a r e n o t e x i s t \\ f_{2} (x) = 5 x^{2} + 6 x + 1 \Rightarrow A_{2} = \frac{\sqrt{36 - 4 (5)}}{5} = \frac{4}{5} \\ f_{3} (x) = x^{2} - 7 x + 6 \Rightarrow A_{3} = \frac{\sqrt{49 - 4 (6)}}{1} = 5 \\ f_{4} (x) = 64 x^{2} + 48 x + 9 \Rightarrow A_{4} = \frac{\sqrt{2304 - 2304}}{64} = 0 \\ ∴ A_{3} > A_{2} > A_{4} \end{array}$