Solving a quadratic equation like (4x^2 – 5x – 12 = 0) involves finding values of (x) that make the equation proper. Quadratic equations are of the form (ax^2 bx c = 0), in which (a), (b), and (c) are constants. The answers to those equations may be determined the usage of numerous techniques, such as factoring, finishing the square, or the use of the quadratic components. In this article, we are able to use the quadratic formulation and also discover an example thru factoring.
Also Read: x2-11x+28=0
Understanding the Quadratic Formula
The quadratic method is given by way of:
[ x = frac-b pm sqrtb^2 – 4ac2a ]
This formulation offers the solutions to any quadratic equation (ax^2 bx c = 0). The term (b^2 – 4ac) is called the discriminant, and it determines the nature of the roots (actual and distinct, actual and same, or complicated).
Solving (4x^2 – 5x – 12 = 0) Using the Quadratic Formula
Identify coefficients: In our equation, (a = four), (b = -five), and (c = -12).
Calculate the discriminant: (b^2 – 4ac = (-5)^2 – 4 times four instances -12).
Substitute inside the formula: Plug those values into the quadratic system and clear up for (x).
Find the 2 solutions: You’ll get values of (x), that are the solutions to the equation.
Example
Let’s solve (4x^2 – 5x – 12 = 0) the usage of the quadratic formula:
Calculate the discriminant:
[ Delta = (-5)^2 – 4 times 4 times -12 = 25 192 = 217 ]
Apply the system:
[ x = frac-(-5) pm sqrt2172 times 4 = frac5 pm sqrt2178 ] This offers us answers:
[ x_1 = frac5 sqrt2178 ]
[ x_2 = frac5 – sqrt2178 ]
Additional Example Through Factoring
Consider a easier quadratic equation, (x^2 – 5x 6 = 0).
Factor the equation: Find two numbers that multiply to ( 6) and upload as much as (-five). These numbers are (-2) and (-3).
Set each element equal to zero:
[ x – 2 = 0 Rightarrow x = 2 ]
[ x – 3 = 0 Rightarrow x = 3 ]
So, the solutions are (x = 2) and (x = three).
Also Read: x*x*x Is Equal To 2
Conclusion
The quadratic components is a reliable approach to locate the roots of any quadratic equation. It’s specially beneficial while the equation isn’t always without problems factorable. Understanding this method and how to practice it’s miles crucial in algebra and prepares students for greater superior mathematical standards. The provided examples illustrate the way to practice this formula and additionally show the factoring approach for less difficult equations.
