Given a quadratic equation, find a value of a such that the quadratic equation would have one root less than -1 and another greater than 1.

Question

I was given this problem:

For what values of $a$ is one of the roots of the equation $$(a^2+a+1)x^2+(2a−3)x+(a−5)=0$$ greater than 1, while the other one is smaller than $1$?

I have no idea how to approach this problem. Can someone possibly help me solve this problem so I can learn how to solve these types of problems?

Answer 1

The hint.

Let $f(x)=(a^2+a+1)x^2+(2a−3)x+a−5.$

Thus, since $$a^2+a+1=\left(a+\frac{1}{2}\right)^2+\frac{3}{4}>0,$$ we need to solve the following inequality: $$f(1)<0.$$

Answer 2

Hint:

It means this polynomial has two real roots and that $1$ separates the roots.

• The discriminant $\Delta=(2a-3)^2-4(a^2+a+1)(a-5)>0$,
• Under this hypothesis, there is a high-school theorem on the separation of roots:

Let $p(x)$ be a quadratic polynomial with two real roots and $\alpha$ be a real number. Then $\alpha$ separates the roots of $p(x)$ if and only if $p(\alpha)$ and the leading coefficient of $p(x)$ have opposite signs.

Here, the leading coefficient is $a^2+a+1$, which is always positive (its roots are the complex cubic roots of unity). So the separating condition is merely $$p(1)=a^2+a+1+2a-3+a-5=a^2+4a-7 <0. $$ Note that if $p(1)<0$, the quadratic polynomial necessarily has tow real roots;, so the first condition is automatically satisfied.