Let's use a trial and improvement method to find this solution.
Step 1. Let's choose x = 8.5
Substituting into the equation:
Step 2. Let's choose x = 8.4
Substituting into the equation:
Step 3. Let's choose x = 8.3
Substituting into the equation:
Since the sign of the equation changes from positive to negative when evaluating from 8.4 to 8.3, then x = 8.3 seems to be a reasonable value. Finally, the solution to 1 decimal place is:
Answer:
An example of a quadratic equation is:
Step-by-step explanation:
This answer follows the standard form for a quadratic equation:
To solve this equation you do the following:
x^2 + 5x + 6 =0
The factors that can be added to get *5* and multiplied to get *6* are 3 & 2.
Therefore the new equation would be:
x^2 +3x + 2x + 6 = 0
<em>group the terms</em>: (x^2 +3x) (+2x + 6) = 0
<em>factorize</em><em>:</em><em> </em>x (x + 3) 2 (x + 3) = 0
<em>therefore</em><em>:</em><em> </em><em>(</em><em>x</em><em> </em><em>+</em><em> </em><em>2</em><em>)</em><em> </em><em>(</em><em>x</em><em> </em><em>+</em><em> </em><em>3</em><em>)</em><em> </em><em>=</em><em>0</em>
<em>x</em><em> </em><em>=</em><em> </em><em>-1</em><em> </em><em> </em><em>x</em><em> </em><em>=</em><em>-3</em>