Question

How do you solve $4{x^2} + 7x < - 3$?

Answer 1

Hint: In this question we are asked to solve the inequality, and the given inequality is quadratic equation of form $a{x^2} + bx + c = 0$, rewrite the middle term as a sum of two terms whose product here it is $a \cdot c = 4 \cdot 3 = 12$ and whose sum is $b = 7$, and simplify the equation till we get the required result.

Complete step by step answer:
Inequalities are mathematical expressions involving the symbols >, <, $ \geqslant $,$ \leqslant $ To solve an inequality means to find a range, or ranges, of values that an unknown x can take and still satisfy the inequality.
Now the give inequality is $4{x^2} + 7x < - 3$,
Now take all terms to one side to make right hand side 0 by adding 3 to both sides, we get,
$ \Rightarrow 4{x^2} + 7x + 3 < - 3 + 3$,
Now simplifying we get,
$ \Rightarrow 4{x^2} + 7x + 3 < 0$,
Now this is a quadratic equation and we will solve by splitting the middle term we get,
Here $a = 4$, $b = 7$ and $c = 3$, so now, $a \cdot c = 4 \cdot 3 = 12$ and $b = 7$,
Now we will factor number 12, i.e., factors of $12 = 1,2,3,4,6,12$,
So, we know that the sum of the number should be equal to $7$, the numbers satisfying the condition are $3$ and $4$ , as there sum will be equal to $7$.
Now rewrite $7$ as$3$ and$4$, then using the equation becomes,
$ \Rightarrow 4{x^2} + 4x + 3x + 3 < 0$,
Now using distributive property, we get
$ \Rightarrow 4{x^2} + 4x + 3x + 3 < 0$,
By grouping the first two terms and last two terms, we get,
$ \Rightarrow \left( {4{x^2} + 4x} \right) + \left( {3x + 3} \right) < 0$,
Now factor out the highest common factor, we get
$ \Rightarrow 4x\left( {x + 1} \right) + 3x\left( {x + 1} \right) < 0$,
Now taking common term in both, we get,
$ \Rightarrow \left( {x + 1} \right)\left( {4x + 3} \right) < 0$,
Now as equating each term, we get,
$ \Rightarrow x + 1 < 0$,
Now simplifying we get,
$ \Rightarrow x < - 1$,
Now equating second term we get,
$ \Rightarrow 4x + 3 < 0$,
Now subtracting 3 on both sides$ \Rightarrow - 1 < x < \dfrac{{ - 3}}{4}$ we get,
$ \Rightarrow 4x + 3 - 3 < 0 - 3$,
Now simplifying we get,
$ \Rightarrow 4x < - 3$,
Now dividing 4 on both sides we get,
$ \Rightarrow \dfrac{{4x}}{4} < \dfrac{{ - 3}}{4}$,
Now simplifying we get,
$ \Rightarrow x < \dfrac{{ - 3}}{4}$,
So, the interval which satisfies $x$ is $ - 1 < x < \dfrac{{ - 3}}{4}$, and this can also be written as $\left( { - 1,\dfrac{{ - 3}}{4}} \right)$.

The solution of the given inequality $4{x^2} + 7x < - 3$ is $\left( { - 1,\dfrac{{ - 3}}{4}} \right)$.

Note: There are three ways to represent solutions to inequalities: an interval, a graph, and an inequality. Because there is usually more than one solution to an inequality, when you check your answer you should check the end point and one other value to check the direction of the inequality. When we work with inequalities, we can usually treat them similarly to but not exactly as we treat equalities.