Introduction:
Polynomial Equations Of Degree Two, Or Quadratic Equations, Can Be Written In The General Form Ax2+Bx+C=0ax^2 + Bx + C = 0ax2+Bx+C=0. 4×2−5x−12=04x^2 – 5x – 12 = 04×2−5x−12=0 Is The Equation. A=4a = 4a=4, B=−5b = -5b=−5, And C=−12c = -12c=−12 Constitute A Quadratic Equation. Finding The Values Of Xxx That Fulfill The Equation Is The First Step In Solving Such Equations. A Quadratic Equation Can Be Solved In A Few Different Ways, Such As Factoring, Completing The Square, And Applying The Quadratic Formula.
Understanding The Quadratic Equation:
Definition:
A Second-Order Polynomial Equation In One Variable Is Called A Quadratic Equation. The Formula In Standard Form Is Ax2+Bx+C=0ax^2 + Bx + C = 0ax2+Bx+C=0, Where A≠0a \Neq 0aʀ=0 And Aaa, Bbb, And Ccc Are Constants.
For Instance:
In The Formula 04x^2 – 5x – 12 = 04×2 – 5x – 12 = 4×2 – 5x – 12 = 0:
- A=4a = 4a=4
- B=-5b = -5b = −5
- C=-12c = -12c = -12
Factoring The Solution:
The Process Of Factoring Entails Representing The Quadratic Equation As A Product Of Binomials.
Procedure Step-By-Step:
Determine The Factors:
Factoring 4×2−5x−124x^2 – 5x – 124×2−5x−12 Is Necessary. In Order To Accomplish This, We Seek Out Two Values That Add Up To Bbb (−5-5−5) And Multiply To A⋅Ca \Cdot Ca⋅C (Which Is 4⋅−12=−484 \Cdot -12 = -484⋅−12=−48).
Following Some Experiments, We Determine That The Integers We Require Are 888 And −6-6−6:
O~8⋅−6 = -488 \Cdot -6 = -488⋅−6 = −48
A Single-Variable, Second-Order Polynomial Equation Is Called A Quadratic Equation. Ax2+Bx+C=0ax^2 + Bx + C = 0ax2+Bx+C=0 Is The Conventional Form, Where A≠0a \Neq 0aʀ=0 And Aaa, Bbb, And Ccc Are Constants.
As An Illustration:
4×2−5x−12=04x^2 – 5x – 12 = 04×2−5x−12=0 Is The Equation In Question.
- A=4a = 4a=4.
- B = -5b = −5
- C = -12c = −12
Factoring An Answer:
In Order To Factor A Quadratic Equation, It Must Be Expressed As A Product Of Binomials.
Procedure In Steps:
Determine Factors:
4×2−5x−124x^2 – 5x – 124×2−5x−12 Must Be Factored. To Do This, We Choose Two Integers That Add Up To Bbb (−5-5−5) And Multiply To A⋅Ca \Cdot Ca⋅C (Which Is 4⋅−12=−484 \Cdot -12 = -484⋅−12=−48).
We Experiment A Little Bit And Discover That The Numbers We Require Are 888 And −6-6−6:
O~8⋅−6 = -488 \Cdot -6 = -488⋅−6 = −48
O~8 + (-6) = 28 + (-6) = 28 + (-6) = 2
Modify The Intermediate Term:
Rewrite −5x-5x−5x With The Following Factors:
4×2 + 3x – 8x – 124×2 + 3x – 8x – 12x^2
Factor And Group Via Grouping:
Sort The Terms And Give Each Group A Factor:
(4×2+3x) + (-8x – 12)(4×2+3x) + (−8x−12) X(4x+3)−4(2x+3)X(4x + 3) – 4(2x + 3)X(4x + 3)-4(2x+3)
Subtract Common Binomial From Total:
Add The Following Factors Together:
(4x+3)(X-4) = 0(4 X + 3)(X – 4) = 0(4 X + 3)(X – 4) = 0
Track Down The Origins:
Put A Zero In Each Factor And Solve:
X + 3 = 0 And X – 4 = 04x+3 = 0 And X – 4 = X = −34 And X = 4x = -\Frac{3}{4} \Quad \Text{And} \Quad X = 4x = −43 And X = 4
Figuring It Out By Finishing The Square:
The Quadratic Equation Must Be Transformed Into A Perfect Square Trinomial In Order To Complete The Square.
Procedure Step-By-Step:
Rephrase The Formula:
Move The Constant Word To The Other Side First:
12 = 124×2 – 5x = 4×2 – 5x = 124×2 – 5x
Divide By The X2x^2×2 Coefficient:
Everything Is Divided By 444.
\Frac{5}{4}X = 3×2−45x=3 – X2−54x=3x^2
Finish The Square:
Add And Subtract (B2a)2\Left(\Frac{B} {2a}\Right) To Finish The Square.^2(2ab)2:
(−542)2=(−5/42)2=(−58)2=2564\Left(\Frac{-\Frac{5}{4}}{2}\Right)^Left(\Frac{-5/4}{2}\Right) = ^2^Left(\Frac{-5}{8}\Right) = ^Two^2 = (−45)\Frac{25}{64}2 = (2−5/4)2 = (8−5)2 = 6425 X2−54x+2564 = 3+2564x^2 – \Frac{5}{4}X + \Frac{25}{64} = 3 + \Frac{25}{64}X2−45x+6425=3+6425 (X−58)2=192+2564\Left(X – \Frac{5}{8}\Right)^2 = \Frac{192 + 25}{64}(X−85)2=64192+25 (X−58)2=21764\Left(X – \Frac{5}{8}\Right)^2 = \Frac{217}{64}(X−85)2=64217
XXX Solution:
Find Xxx By Taking The Square Root Of Both Sides:
X−58=±21764x – \Frac{5}{8} = \Pm \Sqrt{\Frac{217}{64}}X−85=±64217 X−58=±2178x – \Frac{5}{8} = \Pm \Frac{\Sqrt{217}}{8}X−85=±8217 X=5±2178x = \Frac{5 \Pm \Sqrt{217}}{8}X=85±217
Utilizing The Quadratic Formula For Solving:
Any Quadratic Problem Can Be Solved Generally Using The Quadratic Formula: Ax2+Bx+C=0ax^2 + Bx + C = 0ax2+Bx+C=0.
Procedure Step-By-Step:
The Quadratic Formula:
The Equation Is:
X = -B \Pm \Sqrt{B^2 – 4ac}}{2a}X=2a−B±B2−4ac
Alternative Values:
To Solve 4×2−5x−12=04x^2 – 5x – 12 = 04×2−5x−12=0, Replace A=4a = 4a=4, B=−5b = -5b=−5, And C=−12c = -12c=−12 With The Following Values:
X=−(−5)±(−5)2−4⋅4⋅(−12)\Frac{-(-5) \Pm \Sqrt{(-5)^2 – 4 \Cdot 4 \Cdot (-12)}} = 2⋅4x{2 \Cdot 4}X=2⋅4−(−5)±(−5)2−4⋅4⋅(−12) \Frac{5 \Pm \Sqrt{25 + 192}} = X=5±25+1928x{8}\Frac{5 \Pm \Sqrt{217}} = X=85±25+192 X=5±2178x{8}X=85±217
Find The Solution For Xxx:
X = 5 + \Sqrt{217}}{8} \Quad \Text{And} \Quad X = \Frac{5 – \Sqrt{217}}{8} For X = 5+2178 And X = 5−2178.X = 85 + 217 And X = 85 – 217
Summary:
We Investigated Three Approaches To Solve The Quadratic Problem 4×2−5x−12=04x^2 – 5x – 12 = 04×2−5x−12=0: Factoring, Completing The Square, And Applying The Quadratic Formula. The Roots Of The Equation, Which Are Essential For Both Solving And Evaluating Quadratic Functions, Are Provided By Each Approach.
Factoring Formulas: X = -34x = -\Frac{3}{4}X = -43 And X = 4x = 4x = 4
Squaring Up: X=5±2178x = \Frac{5 \Pm \Sqrt{217}}{8}X=85±217
X=5±2178x = \Frac{5 \Pm \Sqrt{217}}{8}X=85±217 Is The Quadratic Formula.
These Solutions Offer A Broader Comprehension Of Polynomial Functions And Show The Adaptability Of Several Approaches To Solving Quadratic Equations.