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Here is the problem

$2x^2+6x+1+k(x^2+2)$, find the condition that must be satisfied by k in order that the expression may be positive for all real values of x.

The quadratic of the form $ax^2+bx+c$ will be positive for all real values of x, if $\displaystyle{\frac{4ac-b^2}{4a} > 0}$.

Therefore, $\displaystyle{\frac{4(2+k)(2k+1)-36}{4(2+k)}>0}$
$\displaystyle{\frac{(k-1)(2k+7)}{(2+k)}}>0$

Considering the 4 ranges, with $k$ values of -14/4, -2 and 1 The inequality is true when $k<-2$ and $k>1$.

However the book answer is only $k>1$. What have I done wrong?

Thank you

Naz
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3 Answers3

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For the quadratic expression $ax^2 + bx+c$ to be greater than zero for all real $x$, two conditions must be satisfied: $a>0$ and the discriminant $b^2 - 4ac<0$

Here $a = k+2, b = 6, c = 2k+1$

The first condition gives $a > 0 \implies k+2 > 0 \implies k > -2$

The second condition gives $b^2 - 4ac < 0 \implies 36 - 4(k+2)(2k+1)<0 \implies 2k^2 + 5k - 7 > 0 \implies (2k+7)(k-1) > 0 \implies k<-\frac 72$ or $k>1$

However, note that $k<-\frac 72$ violates the previously derived condition $k>-2$

Hence the only valid condition is $k>1$

Deepak
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  • I don't understand why discriminant is less than zero. That will give complex roots – Naz Jul 24 '15 at 16:13
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    Precisely. Both roots are complex if the discriminant is less than zero. Which means there are no real roots. Which means there are no intersections between the curve of the quadratic and the $x$-axis. Since the curve is an upright "U"-shape (because of the positive $a$), this guarantees the curve always lies strictly above the $x$-axis, hence the expression is always positive. – Deepak Jul 24 '15 at 16:15
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    @isquared-KeepitReal If, in contrast, you were required to find values of $k$ that would make the expression always negative, you would still need the discriminant to be negative. But in this case the $a$ would also have to be negative. The curve would be an "∩" shape, always lying below the $x$-axis. – Deepak Jul 24 '15 at 16:21
  • This makes sense, thank you. But I am trying to put this into the perspective of what I have learned. And I learned that $\displaystyle{ax^2+bx+c \equiv a(x+\frac{b}{2a})^2 + \frac{4ac-b^2}{4a}}$. From which it follows that $a>0$ and $\displaystyle{\frac{4ac-b^2}{4a}>0}$... There that is what I have missed, the condition on $a$... I will keep in mind your method as well now. Thank you – Naz Jul 24 '15 at 16:37
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    @isquared-KeepitReal What you have learned seems to be completing the square (which is also what Dr Sonnhard Graubner did below). And if you think about it a little, the conditions are equivalent. You still need $a>0$. And in the second expression if $a>0$ then $4a>0$. With a positive denominator, you require a positive numerator to make the whole expression positive, so $4ac -b^2 >0$ or (just rearranging), $b^2 - 4ac < 0$. See? – Deepak Jul 24 '15 at 16:43
  • yeap :) Perfect ! – Naz Jul 24 '15 at 17:24
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The condition $\frac{4ac-b^2}{4a}\gt 0$ is not correct because this includes the case when $a\lt 0$ and $b^2-4ac\gt 0$. (Imagine the graph of a parabola $y=-x^2+x+1$, for example.)

First of all, we need that the coefficient of $x^2$ is positive, i.e. $k+2\gt 0$. (note that $k=-2$ does not satisfy the condition.) Under this condition, you need to have $b^2-4ac\lt 0$, i.e. $9-(2+k)(2k+1)\lt 0$, i.e. $k\lt -\frac 27$ or $k\gt 1$. Thus, we have $k\gt 1$. This is sufficient.

mathlove
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we need a case work. 1. $k=-2$ doesn't work.
2. for $k>-2$ we have $\left(x+\frac{3}{k+2}\right)^2+\frac{2k+1}{k+2}-\frac{9}{(k+2)^2}>0$ for all $x$ The last two summands are $$\frac{(2k+7)(k-1)}{(2+k)^2}$$ this is positive if $k>1$ and the case $k<-\frac{7}{2}$ doesn't exist. 3. the case $k<-2$ is not possible. (Why?)$

Dr. Sonnhard Graubner
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