Why this $12b^2<49ac< \frac{49}{4}b^2$ inequality given ? What is the use of it?

Question

Question :

Let $\alpha$ and $\beta$ be roots of $ax^2+bx+c=0$. Show that $(4\alpha-3\beta)(4\beta-3\alpha)=\frac{49ac-12b^2}{a^2}$. If $12b^2<49ac< \frac{49}{4}b^2$ , then show that $\beta $ lies between $\frac{3}{4}\alpha$ and $\frac{4}{3}\alpha $.

When $\alpha+\beta=-\frac{b}{a} $ and $\alpha \beta = \frac{c}{a} $, I solved a quadratic question where we are asked to show that $$(4\alpha-3\beta)(4\beta-3\alpha)=\frac{49ac-12b^2}{a^2}$$

In the question it is given that $$12b^2<49ac< \frac{49}{4}b^2$$

So by using $12b^2<49ac$ ,

I showed that $\beta $ lies between $\frac{3}{4}\alpha$ and $\frac{4}{3}\alpha $ ,

By arranging $$(\beta-\frac{4}{3}\alpha)(\beta-\frac{3}{4}\alpha)=\frac{12b^2-49ac}{a^2}$$

Why is this part of the inequality given ? What is the use of it ? $$49ac< \frac{49}{4}b^2$$

Answer 1

$$(4\alpha-3\beta)(4\beta-3\alpha)=25\alpha\beta-12(\alpha^2+\beta^2)=49\alpha\beta-2(\alpha+\beta)^2=$$ $$=\frac{49c}{a}-\frac{12b^2}{a^2}=\frac{49ac-12b^2}{a^2}.$$ $49ac<\frac{49b^2}{4}$ says that our equation has two real roots.