Question

How do you solve $4{x^2} - 8x \geqslant 0$ using a sign chart?

Answer 1

First of all, factorize the expression present at the left hand side of the equation (if there is any constant or variable present at right hand side then take it to left hand side then factorize). After factoring, make a table consisting of the factors of the equation and critical values (positive-negative infinity and values of variables you get after comparing the factors with zero). Now check the sign of the function at all critical values and select the interval accordingly that will be the required solution.

Complete step-by-step solution:
In order to solve the given inequality $4{x^2} - 8x \geqslant 0$ using sign chart, we will first factorize the expression of left hand side as follows
$4{x^2} - 8x = 4x(x - 2)$
Now, comparing the factors with zero to get the critical points,
$\Rightarrow 4x = 0\;{\text{and}}\;x - 2 = 0 \\\ \Rightarrow x = 0\;{\text{and}}\;x = 2 \\\$
So $x = 0\;{\text{and}}\;x = 2$ are critical points of the given equation including $\pm \infty$
Now, creating sign chart for the inequality $f(x) = 4{x^2} - 8x \geqslant 0$ as follows

| $- \infty$| $0$| $2$| $\infty$
---|---|---|---|---
$4x$| $-$| $+$| $+$| $+$
$x - 2$| $-$| $-$| $+$| $+$
$f(x)$| $+$| $-$| $+$| $+$

From the sign chart we can see that $f(x)$ is positive from negative infinity to zero and then two to infinity. So we can write the solution for $f(x) = 4{x^2} - 8x \geqslant 0$ as $x \in \left( { - \infty ,\;0} \right] \cup \left[ {2,\;\infty } \right)$
Note: We have included zero and two in the solution because see in the given inequality there is equal sign too which says we should include them as well in the solution. Also when you tackle with fraction type or logarithm type or root type inequalities then do consider their critical points as well as the points where the functions are not defined and exclude them from the set of solutions.