1. ## Quadratic expression

Hey guys,
I'm stuck on this tough question ......

Find the values of a which make x^2 - ax + 1 - 2(a^2) always positive for all real x.

I made the expression equal to 0 ..and then since x is real, the discriminant should be greater than or equal to 0 . I manipulated this inequality to get that x lies between -(2/3) and +(2/3), which is actually the answer given , but I am not satisfied since they said that the the expression should be greater than 0 but I set it equal to 0 .

2. The equation will be 0 only at +- 2/3, so between those, but not including them, the equation will be positive.

3. Hello, Arka!

I don't follow your reasoning, but you seem to have the right answer.

$\text{Find the values of }a\text{ which make }\,x^2 - ax + 1 - 2a^2\,\text{ always positive for all real }x.$

If the expression is positive, the parabola has no x-intercepts.
. . That is, its discriminant is negative.

We have: . $D \;=\; (\text{-}a)^2 - 4(1-2a^2) \:<\:0 \quad\Rightarrow\quad 9a^2-4 \:<\:0$

. . . . . . . . $a^2 \:<\:\tfrac{4}{9} \quad\Rightarrow\quad |a| \:<\:\tfrac{2}{3}$

4. Can we not arrive at the result without relying on co-ordinate geometry for making assumptions like "If the expression is positive, the parabola has no x-intercepts" ?

5. Originally Posted by Arka
Hey guys,
I'm stuck on this tough question ......

Find the values of a which make x^2 - ax + 1 - 2(a^2) always positive for all real x.

I made the expression equal to 0 ..and then since x is real, the discriminant should be greater than or equal to 0 . I manipulated this inequality to get that x lies between -(2/3) and +(2/3), which is actually the answer given , but I am not satisfied since they said that the the expression should be greater than 0 but I set it equal to 0 .
When, in setting f(x)=0, you found the solutions at

$x=\frac{a\pm\sqrt{9a^2-4}}{2}$

then, for "a" real, you have a single solution (graph is tangent to the x-axis) if $9a^2=4$

You have 2 solutions (graph crosses the x-axis) if $9a^2>4$

In both of the above cases, f(x) does not lie ENTIRELY above the x-axis.

f(x) lies entirely above the x-axis if $9a^2<4$

This is because the graph is U-shaped, due to the coefficient of
the highest power of x being positive.

You could also use calculus to examine the minimum value of f(x).

6. Originally Posted by Arka
Hey guys,
I'm stuck on this tough question ......

Find the values of a which make x^2 - ax + 1 - 2(a^2) always positive for all real x.

I made the expression equal to 0 ..and then since x is real, the discriminant should be greater than or equal to 0 . I manipulated this inequality to get that x lies between -(2/3) and +(2/3), which is actually the answer given , but I am not satisfied since they said that the the expression should be greater than 0 but I set it equal to 0 .
Just to throw in my 2 cents one way to do this problem is to complete the square.

$x^2-ax+1-2a^2=x^2-ax+\frac{a^4}{4}-\frac{a^2}{4}+1-2a^2=\left(x-\frac{a}{2} \right)^2+\frac{4-9a^2}{4}$

Now for the quadratic to be positive the term after the square must be positive so we get

$\frac{4-9a^2}{4}> 0 \iff (2-3a)(2+3a)> 0 \iff a \in \left(-\frac{2}{3},\frac{2}{3} \right)$

7. Thanks Guys ,
That really helped !

8. Originally Posted by Soroban
Hello, Arka!

I don't follow your reasoning, but you seem to have the right answer.

If the expression is positive, the parabola has no x-intercepts.
. . That is, its discriminant is negative.

We have: . $D \;=\; (\text{-}a)^2 - 4(1-2a^2) \:<\:0 \quad\Rightarrow\quad 9a^2-4 \:<\:0$

. . . . . . . . $a^2 \:<\:\tfrac{4}{9} \quad\Rightarrow\quad |a| \:<\:\tfrac{2}{3}$

I ma little bit confused in the above expression can u clear about it little more

9. Originally Posted by gramboy
I ma little bit confused in the above expression can u clear about it little more
Be specific. What part confuses you?