**Introduction:**

A Second-Degree Polynomial Equation Of The Type Ax2+Bx+C=0ax^2 + Bx + C = 0ax2+Bx+C=0 Is Called A Quadratic Equation. One Example Of Such An Equation Is 4×2−5x−12=04x^2 – 5x – 12 = 04×2−5x−12=0. We Will Examine Several Approaches For Solving This Quadratic Equation Step-By-Step In This Tutorial.

**Methods To Solve The Quadratic Equation:**

**1.** Calculating

**2.** Finishing Up The Square

**3.** Quadratic Equation

**4.** Let’s Examine Each Technique In More Detail.

**Factoring:**

In Order To Factor A Quadratic Equation, It Must Be Expressed As (Px+Q)(Rx+S)=0.Px + Q + Rx + S = 0 Px + Q + Rx + S = 0. The Equation 4×2−5x−12=04x^2 – 5x – 12 = 04×2−5x−12=0 Can Be Examined.

**Step 1: Multiply The Constant Term (-12) By The Coefficient Of X2x^2×2, Which Is 4:**

4×-12 = −484 \Times -12 = -484×-12 = −48

**Step 2: Determine Two Values That Sum Up To -5 (The Coefficient Of Xxx) And Multiply By -48:**

These Are −8 And 6 So −8×6=−48 And −8+6=−5.Because -8 \Times 6 Equals -48 \Text{ And }, These Integers Are -8 And 6. -8 Plus 6 Equals -5These Are −8 And 6 So −8×6=−48 And −8+6=−5.

**Step 3: Utilizing The Numbers Discovered, Rewrite The Middle Term -5x-5x−5x:**

4×2 – 8x + 6x – 12 = 04x^2 – 8x + 6x – 12 = 04×2 – 8x + 6x – 12 = 0

**Step 4 Grouping Factors:**

4x(-2) + 6(-X) = 04x(-2) + 6(-X) = 04x(-2) + 6(-X) = 0

**Step 5: Factor Out The Common Binomial:**

(4x+6)(X-2)=0(4x – 6) = 0 (4x + 6)(X – 2) = 0

**Step 6: Set Each Factor To Zero And Solve For Xxx:**

X=−324x + 6 = 0 Indicates X = -\Frac{3}{2}. 4x+6=0 ▹X = 2x – 2 = 0 Indicates X = 2x−2=0⟹X=2 Because 4x+6=0⟹X=−23 X−2=0

**Finishing Up The Square:**

**Equation at the end of step1 :**

` ((0 - 2`^{2}x^{2}) - 5x) - 12 = 0

**Pulling out like terms:**

3.1 Pull out like factors :

-4x^{2} – 5x – 12 = -1 • (4x^{2} + 5x + 12)

**Trying to factor by splitting the middle term:**

3.2 Factoring 4x^{2} + 5x + 12

The first term is, 4x^{2} its coefficient is 4 .

The middle term is, +5x its coefficient is 5 .

The last term, “the constant”, is +12

Step-1 : Multiply the coefficient of the first term by the constant 4 • 12 = 48

Step-2 : Find two factors of 48 whose sum equals the coefficient of the middle term, which is 5 .

-48 | + | -1 | = | -49 | ||

-24 | + | -2 | = | -26 | ||

-16 | + | -3 | = | -19 | ||

-12 | + | -4 | = | -16 | ||

-8 | + | -6 | = | -14 | ||

-6 | + | -8 | = | -14 |

For tidiness, printing of 14 lines which failed to find two such factors, was suppressed

Observation : No two such factors can be found !!

Conclusion : Trinomial can not be factored

**Equation at the end of step 3:**

` -4x`^{2} - 5x - 12 = 0

**Parabola, Finding the Vertex:**

4.1 Find the Vertex of y = -4x^{2}-5x-12

Parabolas have a highest or a lowest point called the Vertex . Our parabola opens down and accordingly has a highest point (AKA absolute maximum) . We know this even before plotting “y” because the coefficient of the first term, -4 , is negative (smaller than zero).

Each parabola has a vertical line of symmetry that passes through its vertex. Because of this symmetry, the line of symmetry would, for example, pass through the midpoint of the two x -intercepts (roots or solutions) of the parabola. That is, if the parabola has indeed two real solutions.

Parabolas can model many real life situations, such as the height above ground, of an object thrown upward, after some period of time. The vertex of the parabola can provide us with information, such as the maximum height that object, thrown upwards, can reach. For this reason we want to be able to find the coordinates of the vertex.

For any parabola,Ax^{2}+Bx+C,the x -coordinate of the vertex is given by -B/(2A) . In our case the x coordinate is -0.6250

Plugging into the parabola formula -0.6250 for x we can calculate the y -coordinate :

y = -4.0 * -0.62 * -0.62 – 5.0 * -0.62 – 12.0

or y = -10.438

**Parabola, Graphing Vertex and X-Intercepts:**

Root plot for : y = -4x^{2}-5x-12

Axis of Symmetry (dashed) {x}={-0.62}

Vertex at {x,y} = {-0.62,-10.44}

Function has no real roots

**Solve Quadratic Equation by Completing The Square:**

4.2 Solving -4x^{2}-5x-12 = 0 by Completing The Square .

Multiply both sides of the equation by (-1) to obtain positive coefficient for the first term:

4x^{2}+5x+12 = 0 Divide both sides of the equation by 4 to have 1 as the coefficient of the first term :

x^{2}+(5/4)x+3 = 0

Subtract 3 from both side of the equation :

x^{2}+(5/4)x = -3

Now the clever bit: Take the coefficient of x , which is 5/4 , divide by two, giving 5/8 , and finally square it giving 25/64

Add 25/64 to both sides of the equation :

On the right hand side we have :

-3 + 25/64 or, (-3/1)+(25/64)

The common denominator of the two fractions is 64 Adding (-192/64)+(25/64) gives -167/64

So adding to both sides we finally get :

x^{2}+(5/4)x+(25/64) = -167/64

Adding 25/64 has completed the left hand side into a perfect square :

x^{2}+(5/4)x+(25/64) =

(x+(5/8)) • (x+(5/8)) =

(x+(5/8))^{2}

Things which are equal to the same thing are also equal to one another. Since

x^{2}+(5/4)x+(25/64) = -167/64 and

x^{2}+(5/4)x+(25/64) = (x+(5/8))^{2}

then, according to the law of transitivity,

(x+(5/8))^{2} = -167/64

We’ll refer to this Equation as Eq. #4.2.1

The Square Root Principle says that When two things are equal, their square roots are equal.

Note that the square root of

(x+(5/8))^{2} is

(x+(5/8))^{2/2} =

(x+(5/8))^{1} =

x+(5/8)

Now, applying the Square Root Principle to Eq. #4.2.1 we get:

x+(5/8) = √ -167/64

Subtract 5/8 from both sides to obtain:

x = -5/8 + √ -167/64

In Math, i is called the imaginary unit. It satisfies i^{2} =-1. Both i and -i are the square roots of -1

Since a square root has two values, one positive and the other negative

x^{2} + (5/4)x + 3 = 0

has two solutions:

x = -5/8 + √ 167/64 • i

or

x = -5/8 – √ 167/64 • i

Note that √ 167/64 can be written as

√ 167 / √ 64 which is √ 167 / 8

**Solve Quadratic Equation using the Quadratic Formula:**

4.3 Solving -4x^{2}-5x-12 = 0 by the Quadratic Formula .

According to the Quadratic Formula, x , the solution for Ax^{2}+Bx+C = 0 , where A, B and C are numbers, often called coefficients, is given by :

__ __

– B ± √ B^{2}-4AC

x = ————————

2A

In our case, A = -4

B = -5

C = -12

Accordingly, B^{2} – 4AC =

25 – 192 =

-167

Applying the quadratic formula :

5 ± √ -167

x = ——————

-8

In the set of real numbers, negative numbers do not have square roots. A new set of numbers, called complex, was invented so that negative numbers would have a square root. These numbers are written (a+b*i)

Both i and -i are the square roots of minus 1

Accordingly,√ -167 =

√ 167 • (-1) =

√ 167 • √ -1 =

± √ 167 • i

√ 167 , rounded to 4 decimal digits, is 12.9228

So now we are looking at:

x = ( 5 ± 12.923 i ) / -8

Two imaginary solutions :

x =(5+√-167)/-8=5/-8-i/8√ 167 = -0.6250+1.6154i or: x =(5-√-167)/-8=5/-8+i/8√ 167 = -0.6250-1.6154i

**Two solutions were found:**

- x =(5-√-167)/-8=5/-8+i/8√ 167 = -0.6250-1.6154i
- x =(5+√-167)/-8=5/-8-i/8√ 167 = -0.6250+1.6154i

In Order To Complete The Square, The Quadratic Equation Must Be Converted Into A Perfect Square Trinomial.

**Step 1: Place The Constant Term On The Other Side Of The Equation:**

12 = 124×2 – 5x = 4×2 – 5x = 124×2 – 5x

**Step 2: Divide By X2x^2×2’s Coefficient, Which Equals 4:**

Frac{5}{4}X = 3×2−45x=3 – X2−54x=3x^2

**Step 3: Square The Coefficient Of Xxx And Subtract It From One:**

X2 – 54x + (58)2 – (58)2 = 3x^2 – \Frac{5}{4}X + \Left(\Frac{5}{8}\Right)2-\Left(\Frac{5}{8}\Right)^2 = 3×2 – 45x + (85)2 – (85)2 = 3 X2 – 54x + 2564 = 3 + 2564x^2 + \Frac{25}{64}X2−45x+6425=3+6425 – \Frac{5}{4}X + \Frac{25}{64} = 3

**Step 4: Write As A Perfect Square And Simplify:**

(X-58)2=21764\Left\Frac{217}{64} = (X – \Frac{5}{8}\Right)^2(X-85)2 = 64217

**Step 5: Take The Square Root Of Both Sides To Find The Value Of Xxx:**

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

**Formula For Quadratic Equations:**

X=−B±B2−4ac2ax = \Frac{-B \Pm \Sqrt{B^2 – 4ac}} Is The Quadratic Formula.Ax2+Bx+C=0 Can Be Solved By Using The Formula {2a}X=2a−B±B2−4ac.Ax^2 + Bx + C = 0 Ax2 + Bx + C = 0…

**Recognize Ccc, Bbb, And Aaa:**

A = 4, \, B = -5, \, C = -12a = 4, B = -5, C = -12

**Enter The Following Figures In The Quadratic Formula:**

X=−(−5)±(−5)2−4(4)(−12)\Frac{-(-5) \Pm \Sqrt{(-5)^2 – 4(4)(-12)}}{2(4)} Divides 2(4)X.X=2(4)−(−5)±(−5)2−4(4)(−12) The Equation X=5±25+1928x = \Frac{5 \Pm \Sqrt{25 + 192}}{8}X=85±25+192 X=5±2178x = \Frac{5 \Pm \Sqrt{217}}{8}X=85±217

**Summary:**

We Have Looked At Three Different Approaches To Solving The Quadratic Equation 4×2−5x−12=04x^2 – 5x – 12 = 04×2−5x−12=0 In This Guide. The Quadratic Formula, Factoring, And Completing The Square All Offer A Methodical Way To Approach Solving The Equation. Knowing These Techniques Can Help You Select The Best One For The Particular Quadratic Problem You Are Dealing With.