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  • Find the value of a for which the equation <img alt="x^{2}-3 a x+2a^{2}+a-3=0" src="http://entrancecorner.oncodecogs.com/gif.latex?x%5E%7B2%7D-3%20a%20x&plus;2a%5E%7B2%7D&plus;a-3

Find the value of a for which the equation x^{2}-3 a x+2a^{2}+a-3=0 has real roots and both the roots are less than 1

Option: 1

a\in(-\infty, \frac{ 2 -\sqrt{ 30} } { 6})\cup( \frac{ 2 +\sqrt{ 4 + 24} } { 6},\infty)


Option: 2

a\in(-\infty, 1)\cup( 3,\infty)


Option: 3

a\in(-\infty, -\frac{ 2 } { 3})\cup(1,\frac23)


Option: 4

a\in(-\infty, \frac{2}{3})\cup( 3,\infty)


Answers (1)

best_answer

Location of roots (1) -

let f(x) = ax2 + bx + c where a,b,c is from real number and ‘a’ is non-zero number. Let ? and ? be the solution of the function. And let k is number from real number.  Then:

If both roots of f(x) are less than k then

 

   


 

i) D ≥ 0 (as roots may be equal)

ii) af(k) > 0. As if a < 0 then f(k) < 0. So multiplying two -ve value will give us a positive value, so af(k) > 0 satisfies

iii) \\\mathrm{k > \frac{-b}{2a}\; \;since, \; \frac{-b}{2a}}   will lies between  ? and ?, and ?, ? are less than k so \\\frac{-b}{2a}  will be less than k.

 

-

 

x^{2}-3 a x+2a^{2}+a-3=0

For real roots and both roots less than 1

(i) D>0

\\x^{2}-3 a x+2a^{2}+a-3>0

{\color{Blue} B^2-4AC>0}

\\ 9a^2-4(2a^2+a-3)(1)>0\\ a^2-4a+12>0\\(a-2)^2+8>0\\ a\in R

(ii)-\frac{B}{2A}<1

\\-\frac{-3a}{2}<1 \\\\a<\frac{2}{3}

(iii) af(1)>0

\\1^{2}-2 a (1)+3a^{2}+a-3>0\\ 3a^2-a-2>0\\

a\in(-\infty, \frac{ 1 -\sqrt{ 1+24} } { 6})\cup( \frac{ 1 +\sqrt{ 1 + 24} } { 6},\infty)

a\in(-\infty, -\frac{ 2 } { 3})\cup( 1,\infty)

From (i), (ii), and (iii)

a\in(-\infty, -\frac{2 } { 3})\cup(1,\frac23)

Posted by

seema garhwal

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