Question

How do you solve $2{{x}^{2}}-6x+3\ge 0$ using a sign chart?

Answer 1

We will be using the sign chart for solving the above question. We will first check if the equation has real roots or not using discriminant and we get that ${{b}^{2}}-4ac>0$, that means the equation has 2 real roots (critical values). Then, using these critical values that $x$ can have, we will find the interval in which the given function is equal or greater than 0. The interval in which the given function gives positive value or even a zero is the required interval having the value of $x$, that is, $x\in \left( -\infty ,\dfrac{3-\sqrt{3}}{2} \right]\cup \left[ \dfrac{3+\sqrt{3}}{2},+\infty \right)$.

Complete step by step solution:
According to the given question, we have been given an equation and we have to solve for $x$ such that the function is always positive using sign chart.
Sign chart involves the values $x$ can take up so as to determine the intervals in which the function responds in a way particular way. Based on how function responds in a n interval, we assign the function as increasing, decreasing or 0 in that interval.
We will start by checking if the equation has real roots or not. For this we will use discriminant, which is as follows:
$D={{b}^{2}}-4ac$
We will equate the given equation with the standard quadratic equation, we get,
$2{{x}^{2}}-6x+3=a{{x}^{2}}+bx+c$
$a=2,b=-6,c=3$
Substituting these values in the discriminant, we get,
$\Rightarrow D={{(-6)}^{2}}-4(2)(3)$
$\Rightarrow D=36-24$
$\Rightarrow D=12>0$
Therefore, we have 2 real roots as $D>0$.
Now, to find the value of $x$, we have the quadratic formula, we have,
$x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$
Substituting the values of the variables, we get,
$\Rightarrow x=\dfrac{-(-6)\pm \sqrt{12}}{2(2)}$
$\Rightarrow x=\dfrac{6\pm \sqrt{4\times 3}}{4}$
$\Rightarrow x=\dfrac{2(3)\pm 2\sqrt{3}}{4}$
$\Rightarrow x=\dfrac{3\pm \sqrt{3}}{2}$
Therefore, we have two values of $x$, which are also the critical values of the function
Let’s name them as,
${{x}_{1}}=\dfrac{3-\sqrt{3}}{2}$ and ${{x}_{2}}=\dfrac{3+\sqrt{3}}{2}$
The given expression we have is:
$2{{x}^{2}}-6x+3\ge 0$
Since, we cannot factorize it, we will write a general factor term with respect to the two values we got. Both the terms are positive, so we write the terms as factors as:
$(x-{{x}_{1}})$ and $(x-{{x}_{2}})$, that is,
$2{{x}^{2}}-6x+3\ge 0$
$\Rightarrow (x-{{x}_{1}})(x-{{x}_{2}})\ge 0$
$\Rightarrow f(x)\ge 0$
Now, to find the intervals in which the function increases, we need to have values of $x$, other than the two values we got, so it can range from negative to positive infinity, so $x$ can have the values $(-\infty ,{{x}_{1}},{{x}_{2}},+\infty )$. We now draw the sign chart which is as follows:

Values of $x$	Between $-\infty$ and ${{x}_{1}}$	Between ${{x}_{1}}$ and ${{x}_{2}}$	Between ${{x}_{2}}$ and $+\infty$
$x-{{x}_{1}}$	-ve	+ve	+ve
$x-{{x}_{2}}$	-ve	-ve	+ve
$f(x)$	+ve	-ve	+ve

As we have find the interval in which the function is positive that is greater or equal to zero, therefore, we have,
$x\in \left( -\infty ,\dfrac{3-\sqrt{3}}{2} \right]\cup \left[ \dfrac{3+\sqrt{3}}{2},+\infty \right)$

Note: The critical points (values of $x$) obtained from the function is important to determine the interval in which the function is positive or negative, so the critical points should be calculated correctly. Also, the sign chart should be made carefully. While carrying out the sign analysis, the number taken should be between the defined intervals as per the critical values obtained.