Question

How do you solve $ 3{x^2} - 5x > 8 $ ?

Answer 1

Hint: In order to solve the above equation, subtract both the sides by 8, then use the Quadratic equation formula to solve for the critical points. Now, we will factorize and set factors to zero, so that we can get the values of x and at last we will check the intervals between the critical points.

Complete step-by-step answer:
We are given the inequation (that does not have only ‘=’ sign in it) $ 3{x^2} - 5x > 8 $ .
Subtracting both the sides by 8, and we get:
 $
  3{x^2} - 5x - 8 > 8 - 8 \\
  3{x^2} - 5x - 8 > 0 \;
  $
We obtained a Quadratic equation on the left-hand side:
 $ f(x) = 3{x^2} - 5x - 8 $
Comparing the equation with the standard Quadratic equation $ a{x^2} + bx + c $
 $ a $ becomes $ 3 $
 $ b $ becomes $ - 5 $
And $ c $ becomes $ - 8 $
To find the quadratic factorization we’ll use splitting up the middle term method
So first calculate the product of coefficient of $ {x^2} $ and the constant term which comes to be
 $ = 3 \times - 8 = - 24 $
Now, the second Step is to find the $ 2 $ factors of the number $ - 24 $ such that whether, addition or subtraction of those numbers is equal to the middle term or coefficient of $ x $ and the product of those factors results in the value of constant.
So, if we factorize $ - 24 $ , the answer comes to be $ - 8 $ and $ 3 $ as $ - 8 + 3 = - 5 $ that is the middle term. and $ - 8 \times 3 = - 24 $ which is perfectly equal to the constant value.
Now writing the middle term sum of the factors obtained, so equation $ f(x) $ becomes
 $ f(x) = 3{x^2} + 3x - 8x - 8 $
Now taking common from the first $ 2 $ terms and last $ 2 $ terms
 $ f(x) = 3x(x + 1) - 8(x + 1) $
Finding the common binomial parenthesis, the equation becomes
 $ f(x) = (x + 1)(3x - 8) $
Hence, we have successfully factorized our quadratic equation.
Therefore, the factors are $ \left( {x + 1} \right) $ and $ \left( {3x - 8} \right) $ .
Setting the factors to zero we get the points $ - 1 $ and $ \dfrac{8}{3} $ .
We have two points to mark at the number line $ x = - 1 $ and $ x = \dfrac{8}{3} $ . Mark them and write their signs accordingly and we get:

Since, we know that the left side of the point is always positive and in $ \left( {x + 1} \right)\left( {3x - 8} \right) > 0 $ , $ \left( {3x - 8} \right) $ has the power $ 1 $ which is odd so the sign will change to negative on the left side of $ x = \dfrac{8}{3} $ , similarly for $ \left( {x + 1} \right) $ has also the power $ 1 $ which is odd so again the sign will change to positive.
Overall, according to the equation we were given, $ 3{x^2} - 5x - 8 > 0 $ , means we need the positive value only, so from the number line the positive areas are:
 $ \left( { - \infty , - 1} \right) $ and $ \left( {\dfrac{8}{3},\infty } \right) $ .
Therefore, the factors of the equation $ 3{x^2} - 5x > 8 $ are $ - 1 $ and $ \dfrac{8}{3} $ for the region
 $ \left( { - \infty , - 1} \right) \cup $ $ \left( {\dfrac{8}{3},\infty } \right) $ .
So, the correct answer is “ $ \left( { - \infty , - 1} \right) \cup $ $ \left( {\dfrac{8}{3},\infty } \right) $ ”.

Note:
Always look for the range not only the factors.
The quadratic equation can also be solved using the Quadratic formula.
If we were given only an equation instead of an inequation then we only had to find the factors, not the range.