in my mathematics textbook,i have found one interesting problem and i have one question.textbook asks following problem deduce all possible value of $a$,for which equation $4*x^2-2*x+a=0$ has roots in given interval $(-1;1)$ textbook used following method.it found axis of symmetry $x=-b/2*a$(here $a$ denotes coefficient before $x^2$,so in this case $a=4$) and it got $x=1/4$ because $1/4$ is located in $(-1;1)$,we have to find maximum root which is less then $1$,and minimum one,which is greater then $-1$,also consider all $a$ for which discriminant is non negative and finally we would have set of equation

$1)1-4*a>=0$

$2)(1-\sqrt{1-4*a})/4>-1$

$3)(1+\sqrt{1-4*a})/4<1$

we will get solution $-2<a<=1/4$

my question is what if axis of symmetry is not located in given interval?let us suppose we have equation $5*x^2-17*x+a=0$ here axis of symmetry $x=-b/2*a=17/10=1.7$,but $1.7$ is outside of given interval $(-1;1)$,in this case what i have to do?

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$4x^2-2x+a=0$ have solutions $x=\frac{2\pm \sqrt{4-16a}}{8}$. Since $x\in (-1,1)\implies -1\lt\frac{2\pm \sqrt{4-16a}}{8}\lt 1\implies -8\lt {2\pm \sqrt{4-16a}}\lt 8\implies -10\lt {\pm \sqrt{4-16a}}\lt 6$. Solve for these inequalities, you will get the required values for $a$.
for $ax^2+bx+c=0\implies x^2+bx/a+c/a=0\implies x^2+2(b/a)x+b^2/(4a^2)+c-b^2/(4a^2)=0\implies (x+b/(2a))^2=\frac{b^2-4ac}{4a^2}\implies |x+b/(2a)|=\frac{\sqrt D}{2a}$. Thus if take axis of symmetry $x=-b/2a$, it reduces to $|y|=\frac{\sqrt D}{2a}$ which is an easier form to solve your problem. – Aang Jul 13 '12 at 7:53