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The values of a for which both roots of quadratic equation x^2-2ax-a lie in the interval (-2,2), are

Option: 1

a\in(-\infty,-1]\cup[0,\infty)


Option: 2

a\in(-\infty,-1)\cup(0,\infty)


Option: 3

a\in(-\frac{4}{3},-1]\cup[0,\frac{5}{4})


Option: 4

a\in[-\frac{4}{3},-1]\cup[0,\frac{5}{4}]


Answers (1)

best_answer

Required conditions are

i) D ≥ 0 (as roots may be equal)

ii) \\\mathrm{af(k_1) > 0 \; and \; af(k_2) > 0\;} 

iii) \\\mathrm{ k_1 <\frac{-b}{2a} < k_2,\; where\; \alpha \leq \beta \; and \; k_1 < k_2}

 

Now,

x^2-2ax-a

If both the roots lie in the interval (-2,2) then, 

\\(i)\;\;\;D\geq0\\ (ii) \;\;af(-2)>0 \\(iii)\;af(2)>0 \\(iv) -2 < \frac{-b}{2a} < 2

 

i) 

\\D\geq0\\ 4a^2+4a\geq0\\a(a+1)\geq0\\ a\in(-\infty,-1]\cup[0,\infty)

ii)

\\af(-2)>0\\ (-2)^2-2(-2)a-a>0\\ 4+3a>0\\ a>-\frac{4}{3}

iii)

\\af(2)>0\\ (2)^2-2(2)a-a>0\\ 4-5a>0\\ a<\frac{5}{4}

iv)

\\-2 < \frac{-b}{2a} < 2\\\\-2 < \frac{2a}{2}< 2\\\\-2<a<2

 

From (i), (ii), (iii) and (iv), intersection is

a\in(-\frac{4}{3},-1]\cup[0,\frac{5}{4})

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manish

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