Quadratic EquationBY- Utkarsh Sharma

2x2 + 5x + 3 = 0 =In this

one a=2, b=5 and c=3

x2 − 3x = 0= 

•This one is a little more tricky: Where is a? Well a=1, and we don't usually write

"1x2"•b = -3

•And where is c? Well c=0, so is not shown.

5x − 3 = 0

 =

Oops! This one is not a quadratic equation: it is

missing x2 (in other words a=0, which means it can't be quadratic)

So far we have seen the "Standard Form" of a Quadratic Equation:

But sometimes a quadratic equation doesn't look like that..! Here are some examples..

In disguise In Standard Form a, b and c

x2 = 3x -1 Move all terms to left hand side

x2 - 3x + 1 = 0 a=1, b=-3, c=1

2(w2 - 2w) = 5 Expand (undo the brackets), and move 5 to left

2w2 - 4w - 5 = 0 a=2, b=-4, c=-5

z(z-1) = 3 Expand, and move 3 to left z2 - z - 3 = 0 a=1, b=-1, c=-3

5 + 1/x - 1/x2 = 0 Multiply by x2 5x2 + x - 1 = 0 a=5, b=1, c=-1

Factor method…………

…Step 1: Find two numbers that multiply to give ac, and add to give b.

• Here .. a = 2 ,, …. b = 7…, c= 3• ac is 2×3 = 6 and b is 7• So we want two numbers that multiply together to make 6, and

add up to 7• In fact 6 and 1 do that (6×1=6, and 6+1=7)• How do we find 6 and 1?• With the help of  the factors of ac=6, and then try adding some

to get b=7.• Factors of 6 include 1, 2, 3 and 6.• Aha! 1 and 6 add to 7, and 6×1=6.

Step 2:  Rewrite the middle with those numbers:So we..............Rewrite 7x with 6x and 1x:2x2 + 6x + x + 3

Step 3: : Factor the first two and last two terms separately:The first two terms 2x2 + 6x factor into 2x(x+3)The last two terms x+3 don't actually change in this caseSo we get:2x(x+3) + (x+3)

Step 4: If we've done this correctly, our two new terms should have a clearly visible common factor.In this case we can see that (x+3) is common to both termsSo we can now rewrite it like this:2x(x+3) + (x+3) = (2x+1)(x+3) = 2x2 +7x + 3(hence it’s correct)

•

x2 + bx + (b/2)2=(x+b/2)2In Algebra it looks like this:by adding (b/2)2 we can complete the square. And (x+b/2)2 has x only once, which is easier to use.

A Quadratic Equation looks like this:And it can be solved using the Quadratic Formula:

Move the number term to the right side of the equation: x2 + 4x = -1

Complete the square on the left side of the equation and balance this by adding the same number to the right side of the equation.

(b/2)2 = (4/2)2 = 22 = 4 x2 + 4x + 4 = -1 + 4 (x + 2)2 = 3 Take the square root on both sides of the

equation: x + 2 = ±√3 = ±1.73 Subtract 2 from both sides: x = ±1.73 – 2 = -3.73 or -0.27

• A Quadratic Equationlooks like this:

And it can be solved using the Quadratic Formula:

The "±" means you need to do a plus AND a minus, so there are normally TWO solutions !The blue part (b2 - 4ac) is called the "discriminant“or ‘D’, because it can "discriminate" between the possible types of answer.

D = 0 , then root’s would be equal.D > or = 0, Then root’s would be real.D < 0 ,Then root’s would be unreal or imaginary or there would be complex solutions .

• By using quadratic formula –• Substitute a=6, b=5 and c=−6 into the formula:• x = (−b ± √[b2 − 4ac]) / 2a• x = (−5 ± √[52 − 4×6×(−6)]) / 2×6• = (−5 ± √[25 + 144]) / 12• = (−5 ± √169) / 12• = (−5 ± 13) / 12• So the two roots are: • x = (-5 + 13) / 12 = 8/12 = 2/3,• x = (-5 − 13) / 12 = −18/12 = −3/2

Explore the Quadratic Equation -

Graphical Method

Explore the Quadratic Equation -

Graphical Method

x x2 - 4 y-3 (-3)2 - 4 5-2 (-2)2 - 4 0-1 (-1)2 - 4 -30 (0)2 - 4 -41 (1)2 - 4 -32 (2)2 - 4 03 (3)2 - 4 5

The height starts at 3 m:   3

It travels upwards at 14 meters per second (14 m/s):   14t

Gravity pulls it down, changing its speed by about 5 m/s per second (5 m/s2):  −5t2

(Note for the enthusiastic: the -5t2 is simplified from -(½)at2 with a=9.81 m/s2)   

Multiply all terms by −1 to make it easier: 5t2 − 14t − 3 = 0Now our job is to factor it.

We will use the "Find two numbers that multiply to give a×c, and add togive b" method in Factoring Quadratics.

a×c = −15, and b = −14.The positive factors of −15 are 1, 3, 5, 15, and one of the factors 

has to be negative.By trying a few combinations we find that −15 and 1 work 

(−15×1 = −15, and −15+1 = −14)

Rewrite middle with -15 and 1:   5t2 − 15t + t − 3 = 0

Factor first two and last two:   5t(t − 3) + 1(t − 3) = 0

Common Factor is (t - 3):   (5t + 1)(t − 3) = 0

     

And the two solutions are:   5t + 1 = 0 or t − 3 = 0

    t = −0.2  or  t = 3