Quadratic Equations

Quadratic Equations

Standard Form The Standard Form of Form  of a Quadratic Equation looks like this:

 +  +  = 0,  ≠ 0 2

Where a ,b ,c are real numbers x is variable or Unknown.

 ++3=0

For example 7

2

Note:If a=0, it will not be a quadratic equation. The name Quadratic comes from "quad" meaning square, because the variable gets squared (like x2).It is also called an "Equation of Degree 2" (because of the "2" "2 " on the x)

Pure Quadratic Equations The Pure Form of a Quadratic Equation looks like this:

b=0 in Standard quadratic Equation Equation  +  +  = 0  +  = 0, b=0 For example  − 1 6 = 0 , 4 = 7 are pure form of quadratic equation. 2

2

2

2

How To Solve Quadratic Equation ? There are three Methods to solve a Quadratic Equation. 1. Factorization 2. Completing the Square

Ali Hassan Tahir M.Sc. Computational Physics Email [email protected]

Page 1

Quadratic Equations 3. By using Quadratic Formula

Exercise 1.1

th

10 -Class

1. Write the following Quadratic Equations in the standard form and point out pure Quadratic Equations.

 + − = −7 Soln.  + 7 − 3 = −7  + 4 − 21 = −7  + 4 − 21 + 7 = 0  + 4 − 14 = 0 (Standard form) =>

(i)

2

2

2

+ −  =   

(ii)

Soln.

+ −  =   

Taking L.C.M of 3,7 (which is 21) and multiplying on both side We get

 − 21   = 21(1) 7 + 4 − 3 = 21 7 + 28 − 3 = 21 7 + 28 − 3 − 21 = 0 => 7 − 3 + 7 = 0 2 +4

21

3

7

2

2 2

2

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(Standard form)

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Quadratic Equations (iii)



+ =  +  +

  +1  +1 +  = 6

Sol.

Multiply both sides by (x+1)(x). (L.C.M Of (x+1) and (x)) We get

 + 1()   +  + 1()   = 6 + 1()  +  + 1 + 1 = 6( + )  +  + 2 + 1 − 6 − 6 = 0 −4 − 4 + 1 = 0 Taking (-) common we get => 4 + 4 − 1 = 0 (Standard form) + − − +  =  (iv) −   − Sol. − −  + 4 = 0 +1

+1

2

2

2

2

2

2

2

+4

2

2

Multiply both sides by (x-2)(x). (L.C.M Of (x-2) and (x)) We get

 − 2 − −  − 2 − + 4 − 2 = 0 − 2()  + 4 −  − 2 − 2 + 4 − 2 = 0  + 4 −  − 4 + 4 + 4 − 8 = 0 4 − 4 = 0 Taking (4) common we get =>  − 1 = 0 (Pure quadratic Form) + − − =  (v) +  +4

2

2

2

2

2

2

2

2

Do your self Same As Part (iv) Ans. (Pure quadratic Form)

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Page 3

Quadratic Equations (vi) Sol.

+ + + =  + +  + + + =  + + 

Multiply both sides by (x+2)(x+3)(12). (L.C.M Of (x+2) , (x+3) and (12)) We get

 + 2 + 312  +  + 2 + 312  =  + 2 + 312   + 312( + 1 ) +  + 212 + 2 =  + 2 + 325 12x + 3( x + 1 ) + 12x + 2x + 2 = 25x + 2x + 3 12( + 4 + 3 ) + 1 2 ( + 4 + 4) = 25( + 5 + 6) Simplifying we get (12 + 12 − 25) + 48 + 48 − 150 + (38 + 48 − 150) = 0 − − 29 − 66 = 0 Taking (-) common we get =>  + 29 + 66 = 0 (Standard form) 2

+1

+2

25

+2

+3

12

2

2

2

2

2

What is factorization?

Let us consider a simple example (Numbers) 12 = 3 × 4 i.e., 12 is product of 3 and 4. 3 and 4 are called factors or divisors of 12 12 is also equal to 2 × 6. Similarly

an Algebraic expression can be Factorized .

For Example

 + 4 + 3 = 0 have Factors as 2

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Factorization Expressing polynomials as  product of other polynomials that cannot be further  factorized is called Factorization

Page 4

Quadratic Equations Q2. Solve by Factorization.

 − − 20 = 0 Sol.  −  − 20 = 0  − 5 + 4 − 20 = 0  − 5 + 4 − 5 = 0  − 5 + 4 = 0  − 5 = 0 or  + 4 = 0 =>  = 5 or  = −4 2

(i).

Some tips For Factorization.

 +  +  = 0

2

2

Solution Set = {5,-4}

 = (− 5) Sol. 3 = ( − 5) 3 − + 5 = 0 2 + 5 = 0 (ii) 3

2

In this Equation. 

If c is positive, then the factors you're looking



for are either both positive or else both negative. (factors that you're looking for add to b.)

2

2

2

 If c is negative, then the

2

factors you're looking for are of alternating signs; that is, one is negative and one is positive.

2

Taking y common, we get

2 + 5 = 0  = 0  2 + 5 = 0  = 0  = − 5 2

5

Solution Set = {0,- } 2

− 32 = 17 Sol. 4 − 32 = 17 4 − 32 − 17 = 0 Taking (-) common 17 + 32 − 4 = 0 17 + 34 − 2 − 4 = 0 2

(iii) 4

2

2

2

2

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Page 5

Quadratic Equations

 + 2 − 2 + 4 = 0 x + 217x − 2 = 0 Thus, x + 2 = 0 or 17x − 2 = 0 x = −2, 17x = 2 = 17

2

17

Solution Set = {-2,

2 17

}

 − 11 = 152 Sol. − 11 = 152  − 11 − 152 = 0  − 19 + 8 − 152 = 0  − 19 + 8( − 19) = 0  − 19( + 8) = 0 ℎ ,  − 19 = 0   + 8 = 0 x = 19, x = −8 2

(iv)

2

2

2

Solution Set = {19,-8}

+1 +  = 25   +1 12 +1 +  = 25 Sol.  +1 12

(v)

  × 12 ( + 1 ) +  × 12( + 1 ) = × 12 ( + 1)   12 + 1 + 1 = 12  = 25 ( + 1) 12 + 2 + 1 + 12 = 25( +  ) 12 + 24 + 1 2 + 1 2 = 25 + 25

Multiplying both sides by 12 ( + 1) +1

25

+1

2

2

12

2

2

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2

2

Page 6

Quadratic Equations

 + 24 + 1 2 + 1 2 − 25 − 25 = 0 12 + 12 − 25 + 24 − 25 + 12 = 0 − − + 12 = 0 Taking (-) common we get   +  − 12 = 0  + 4 − 3 − 12 = 0  + 4 − 3 + 4 = 0  + 4 − 3 = 0 ℎ ,  + 4 = 0   − 3 = 0  = −4 ,  = 3 12

2

2

2

2

2

2

2

2

2

Solution Set = {-4,3} (vi)

−9 = −3 − −4 2

Sol.

1

2

=

1

1

− 9 −3 −4−(x −3) 2 −9 = −3(− 4) 2 −4− +3 −9 = − 3(− 4) 2 −1 −9 = − 3(− 4)

− − ( .  .    . .  ) 1

4

Now by cross multiplication, we have

 − 3 − 4 = −1( − 9) 2 − 7 + 12 = − + 9 2 − 14 + 2 4 +  − 9 = 0 2 − 13 + 15 = 0 2

2

2

2

Now Factorizing

 − 10 − 3 + 15 = 0 2  − 5 − 3 − 5 = 0 2

2

Ali Hassan Tahir M.Sc. Computational Physics Email [email protected]

Page 7

Quadratic Equations

 − 52 − 3 = 0 ℎ  − 5 = 0  2 − 3 = 0 =5 , = 3 2

Solution Set =

3

{5, } 2

Now we will learn completing Square Technique to solve quadratic Equation. Derivation of Quadratic Formula using completing square. let's go

 +  +  = 0 2

 

(1)

Step-1 Coefficient of

  should be 1. If it is not 1 make it 1 by dividing. 2

“

”

Dividing Above Equation by a  ,We get

 

2

+

 +  = 0   

 

Step-2 Put Constant term Over the Other side (i.e. .)



2

+

 = −   

Step-3 Take half of Coefficient of x-term, and square it and add on both sides.



2

+

  +   2 = −  +   2  2  2

Step-4 Convert the left-hand side to square forms and simplify on the right-hand side.

 + 2    +    = −  +   +   = −   2

2

2

2

2

4

2

+

2

4 2

2

4 2

Step-4 Convert left-hand side in square  form using this

 + 2 +  =  +  2

2

2

Step-5 Square-root both sides, remembering to put the "±" on the right. Ali Hassan Tahir M.Sc. Computational Physics Email [email protected]

Page 8

Quadratic Equations

  +   = ±  −    +  = ±   −  2

4

+ 2

4 2

2

2

2

4

2

Step-6 Solve for "x =", and simplify as necessary.

 = −   −  ,   = −−  −  2

+

2

4

2

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4

2

Page 9

Quadratic Equations

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Page 10