Mathematics - Quadratic Equation Question with Solution | TestHub

Given equation is $x^2+9y^2-4x+3=0$.....(i)

Or, $x^2-4x+9y^2+3=0$.

Since $\mathbf{x}$ is real

$$\therefore (-4{)}^2-4\left(9y^2+3\right)\ge 0$$

Or, $16-4\left(9y^2+3\right)\ge 0$ or, $4-9y^2-3\ge 0$

Or, $9y^2-1\le 0$ or, $9y^2\le 1$ or, $y^2\le \frac{1}{9}$

Now $y^2\le \frac{1}{9}\iff -\frac{1}{3}\le y\le \frac{1}{3}$ ...........(ii)

Equation (i) can also be written as

$9y^2+0y+x^2-4x+3=0.\dots \dots \left(iii\right)$

Since $y$ is real $\therefore 0^2-4.9\left(x^2-4x+3\right)\ge 0$

Or, $\begin{array}{rlr}& x^2-4x+3\le 0& \Rightarrow x\in [1,3]\end{array}$