JEE MAIN - Mathematics (2020 - 5th September Morning Slot - No. 20)
The product of the roots of the
equation 9x2 - 18|x| + 5 = 0 is :
equation 9x2 - 18|x| + 5 = 0 is :
$${{5} \over {9}}$$
$${{5} \over {27}}$$
$${{25} \over {81}}$$
$${{25} \over {9}}$$
Explanation
$$9{x^2} - 18\left| x \right| + 5 = 0$$
$$ \Rightarrow $$ $$9{x^2} - 15\left| x \right| - 3\left| x \right| + 5 = 0$$ ($$ \because $$ x2 = $${\left| x \right|^2}$$)
$$ \Rightarrow $$ $$3\left| x \right|(3\left| x \right| - 5) - (3\left| x \right| - 5) = 0$$
$$ \Rightarrow $$$$\left| x \right| = {1 \over 3},\,{5 \over 3}$$
$$ \Rightarrow $$ $$x = \pm {1 \over 3}, \pm \,{5 \over 3}$$
$$ \therefore $$ Product of roots
= $$\left( \frac{1}{3} \right) \left( -\frac{1}{3} \right) \left( \frac{5}{3} \right) \left( -\frac{5}{3} \right) $$ = $${{25} \over {81}}$$
$$ \Rightarrow $$ $$9{x^2} - 15\left| x \right| - 3\left| x \right| + 5 = 0$$ ($$ \because $$ x2 = $${\left| x \right|^2}$$)
$$ \Rightarrow $$ $$3\left| x \right|(3\left| x \right| - 5) - (3\left| x \right| - 5) = 0$$
$$ \Rightarrow $$$$\left| x \right| = {1 \over 3},\,{5 \over 3}$$
$$ \Rightarrow $$ $$x = \pm {1 \over 3}, \pm \,{5 \over 3}$$
$$ \therefore $$ Product of roots
= $$\left( \frac{1}{3} \right) \left( -\frac{1}{3} \right) \left( \frac{5}{3} \right) \left( -\frac{5}{3} \right) $$ = $${{25} \over {81}}$$
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