Introduction:
Fundamental To Algebra, Quadratic Equations Are Used To Describe A Wide Range Of Real-World Issues. Any Equation That Can Be Expressed In The Form Ax2+Bx+C=0ax^2 + Bx + C = 0ax2+Bx+C=0, Where A≠0a \Neq 0aʀ=0 And Aaa, Bbb, And Ccc Are Constants, Is A Quadratic Equation. Using Factoring, The Quadratic Formula, And Completing The Square, We Will Solve The Quadratic Equation 4×2−5x−12=04x^2 – 5x – 12 = 04×2−5x−12=0 In This Tutorial.
Understanding The Quadratic Equation:
What Is A Quadratic Equation?
As A Second-Degree Polynomial Equation, A Quadratic Equation Has A Variable Xxx With A Maximum Power Of 2. The Typical Format Is:
Ax^2 + Bx + C = 0ax2+Bx+C=0 Ax2+Bx+C=0
The Equation 4×2−5x−12=04x^2 – 5x – 12 = 04×2−5x−12=0 Can Be Examined.
- The Coefficient Of X2x^2×2 Is A=4a = 4a=4.
- B=-5b = -5b = −5 (The Xxx Coefficient)
- The Constant Term, C=-12c=-12,
Factoring The Solution:
Quadratic equations form a fundamental part of algebra and have significant applications in various fields including physics, engineering, and economics. In this article, we will discuss on how to solve the quadratic equation 4x^2 – 5x – 12 = 0 and discuss what exactly are Quadratic equations and different methods to solve Quadratic equations.
What are Quadratic Equations?
Quadratic equations are math problems where the variable is squared (raised to the power of two). They are very important in algebra and are used in subjects like physics, engineering, and economics. Typical the Quadratic Equations looks like ax^2 + bx + c = 0 where a, b, and c are constants, and x is the variable.
Here in this article we will explain different methods to solve Quadratic Equations and also explain how to solve 4x^2 – 5x – 12 = 0.
For this specific equation 4x^2 – 5x – 12 = 0:
a = 4, b = -5, c = -12
Methods to Solve the Quadratic Equation 4x^2 – 5x – 12 = 0:
There are several methods to solve quadratic equations, including:
1. Factoring
2. Completing the Square
3. Using the Quadratic Formula
4. Graphical Method
Let’s apply these methods to solve 4x^2 – 5x – 12 = 0.
Factoring Method to Solve 4x^2 – 5x – 12 = 0:
Factoring involves expressing the quadratic equation as a product of two binomials. However, not all quadratic equations can be factored easily. Let’s see if this method applies to our equation.
We look for two numbers that multiply to a * c = 4 * -12 = -48 and add to b = -5. These numbers are -8 and 6.
Thus, we rewrite the middle term -5x using -8 and 6:
4x^2 – 8x + 6x – 12 = 0
Next, we factor by grouping:
4x(x – 2) + 6(x – 2) = 0
(4x + 6)(x – 2) = 0
This gives us the factors (4x + 6) and (x – 2). Setting each factor to zero provides the solutions:
4x + 6 = 0 => x = -3/2
x – 2 = 0 => x = 2
Completing the Square Method to Solve 4x^2 – 5x – 12 = 0:
Completing the square involves rewriting the quadratic equation in the form (x – p)^2 = q. Here’s the process:
Start with the original equation:
4x^2 – 5x – 12 = 0
Divide through by 4 to simplify:
x^2 – 5/4x – 3 = 0
Move the constant term to the right side:
x^2 – 5/4x = 3
Add the square of half the coefficient of x to both sides:
x^2 – 5/4x + (5/8)^2 = 3 + (5/8)^2
x^2 – 5/4x + 25/64 = 3 + 25/64
(x – 5/8)^2 = 217/64
Take the square root of both sides:
x – 5/8 = ±√(217/64)
x = 5/8 ± √(217)/8
Simplifying further:
x = (5 ± √(217))/8
Therefore, x can be either 2.466 or −1.216-.
These are the roots of the equation, although in this particular case, completing the square leads to a more complex expression.
Quadratic Formula Method to Solve 4x^2 – 5x – 12 = 0:
The quadratic formula is a universal method to solve any quadratic equation:
x = (-b ± √(b^2 – 4ac)) / 2a
For our equation 4x^2 – 5x – 12 = 0:
a = 4, b = -5, c = -12
Plugging these values into the formula:
x = (5 ± √(25 + 192)) / 8
x = (5 ± √(217)) / 8
Thus, the solutions are:
x = (5 + √(217)) / 8
x = (5 – √(217)) / 8.
Therefore, x can be either 2.466 or −1.216-.
Graphical Method to Solve 4x^2 – 5x – 12 = 0:
Graphing the quadratic equation provides a visual representation of its solutions. The graph of y = 4x^2 – 5x – 12 is a parabola. The x-intercepts of this parabola correspond to the solutions of the equation 4x^2 – 5x – 12 = 0.
To sketch the graph:
1. Identify the vertex using x = -b/2a:
x = 5/8
2. Calculate y at the vertex:
y = 4(5/8)^2 – 5(5/8) – 12
y = 4(25/64) – 25/8 – 12
y = 100/64 – 200/64 – 768/64
y = (100 – 200 – 768) / 64 = -13.5625
3. Plot the vertex (5/8, -13.5625) and the x-intercepts x = -3/2 and x = 2.
The parabola opens upwards (since a = 4 > 0) and crosses the x-axis at the solutions we found.
Procedure Step-By-Step:
Rephrase The Formula: Let’s Begin With The Equation:
04x^2 – 5x – 12 = 04×2 – 5x – 12 = 4×2 – 5x – 12 = 0
Reduce The Quadratic Formula:
O We Must Choose Two Integers That Add Up To Bbb (Which Is −5-5−5) And Multiply To A⋅Ca \Cdot Ca⋅C (Which Is 4⋅−12=−484 \Cdot -12 = -484⋅−12=−48).
O 666 And −8-8−8 Are These Two Numerals Because:
−8⋅6 = -48−8⋅6 = -48 −8+6 = -5-8 + 6 = -5−8+6 = −5
Modify The Intermediate Term:
O Divide −5x-5x−5x Into −8x+6x-8x + 6x−8x+6x, The Middle Term:
4×2 – 8x + 6x – 12 = 04x^2 – 8x + 6x – 12 = 04×2 – 8x + 6x – 12 = 0
Factor Based On Grouping:
O Group The Terms And Give Each Group A Factor.
(4×2−8x) + (6x−12) = 04x^2 – 8x) + (6x – 12) = 0 (4x^2 – 8x) + (6x – 12) = 0 4x(-2) + 6(-X) = 04x(-2) + 6(-X) = 04x(-2) + 6(-X) = 0
O Divide Both Sides Of The Common Binomial By (X-2)(X – 2)(X-2):
(4x+6)(X-2)=0(4x – 6) = 0 (4x + 6)(X – 2) = 0
Find The Solution For Xxx:
O Solve By Setting All Of The Factors To Zero:
4 X + 6 = 4 X + 6 = 0 X−2=0x – 2 = 0x−2=0
O Resolving These Formulas:
4 X + 6 = 4 X + 6 = 0 4x = -64x = -64x = −6 X = −32x = -\Frac{3}{2}X = −23 X−2=0x – 2 = 0x−2=0 X=2x = 2x=2
O X=2x = 2x=2 And X=−32x = -\Frac{3}{2}X=−23 Are The Solutions.
Utilizing The Quadratic Formula For Solving:
The Formula For Quadratic Equations:
The Roots Of Any Quadratic Equation Can Be Found Using The Quadratic Formula, Which Is Provided By:
X = -B \Pm \Sqrt{B^2 – 4ac}}{2a}X=2a−B±B2−4ac
With Regard To The Formula 4×2−5x−12=04x^2 – 5x – 12 = 04×2−5x−12=0:
- A=4a = 4a=4
- B=-5b = -5b = −5
- C=-12c = -12c = -12
Procedure Step-By-Step:
Determine The Discriminant:
Δ=B2−4acdelta Is Equal To B^2 – 4ac.Δ=B2−4ac Δ=(−5)2−4⋅4⋅(−12)\(-5)^2 – 4 \Cdot 4 \Cdot (-12) = Deltaδ=(−5)2−4⋅4⋅(−12) Δ=25+192\Delta Is Equal To 25 Plus 192.Δ=25+192 Δ=217\Delta = 217Δ = 217
Utilize The Quadratic Equation:
X = -(-5) \Pm \Sqrt{217}}{2 \Cdot 4} = −(−5)±2172⋅4xx=5±2178x = \Frac{5 \Pm \Sqrt{217}}{8} X=2⋅4−(−5)±217x Equals 85±217
O These Are The Two Options:
\Frac{5 + \Sqrt{217}}{8} = X=5+2178xx Equals 85 Plus 217 \Frac{5 – \Sqrt{217}}{8} = X=5−2178xx = 85 – 217
Using The Square To Complete The Solving:
Procedure Step-By-Step:
Put The Equation Back In Standard Form:
12 = 124×2 – 5x = 4×2 – 5x = 124×2 – 5x
Split All Terms By Aaa, Or Four:
\Frac{5}{4}X = 3×2−45x=3 – X2−54x=3x^2
Finish The Square:
O Within The Equation, Add And Subtract (B2)2\Left(\Frac{B}{2}\Right)^2(2b)2:
X2−54x+(58)2=3 + (58)2x^2 – \Left(\Frac{5}{8}\Right) + \Frac{5}{4}X3 + \Left(\Frac{5}{8}\Right) = ^2^2×2−45x+(85)2=3 + (85)2.
(58)2\Left(\Frac{5}{8}\Right)^2(85)2 Must Be Calculated.
The Formula (58)2=2564\Left(\Frac{5}{8}\Right)^2 = \Frac{25}{64}(85)2=6425
O Replace Once More In The Formula:
(X-58)2=3+2564\Left(X – \Frac{5}{8}\Right)^2 = 3 + \Frac{25}{64}(X−85)2=3+6425 (X−58)2=19264+2564\Left(X – \Right\Frac{5}{8})^2 = (X-85)2 + (192}{64})²= 64192 + 6425 (X-58)2 = 21764\Left\Frac{217}{64}(X−85) = (X – \Frac{5}{8}\Right)^264217 ÷ 2 =
XXX Solution:
X−58=±2178x – \Frac{5}{8} = \Pm \Frac{\Sqrt{217}}{8}X−85=±8217 X=58±2178x = \Frac{5}{8} \Pm \Frac{\Sqrt{217}}{8}X=85±8217
O These Are The Fixes:
\Frac{5 + \Sqrt{217}}{8} = X=5+2178xx Equals 85 Plus 217 \Frac{5 – \Sqrt{217}}{8} = X=5−2178xx = 85 – 217
Final Thoughts:
There Are Several Methods For Solving Quadratic Equations, And Each Is Helpful In A Particular Situation. We Determined The Answers For The Equation 4×2−5x−12=04x^2 – 5x – 12 = 04×2−5x−12=0 Using:
- Factoring: X=−32x = -\Frac{3}{2}X=−23 And X=2x = 2x=2.
- Quadratic Formula: X=5±2178x = \Frac{5 \Pm \Sqrt{217}}{8}X=85±217.
- Squaring Up: X=5±2178x = \Frac{5 \Pm \Sqrt{217}}{8}X=85±217
Gaining Knowledge Of These Techniques Will Enable You To Solve A Wide Range Of Quadratic Problems Efficiently.
