Easy way of solving quadratic equations
Step 1: multiply the
coefficient of first term with the last term.
Step 2: Find the 2 factors of 6
that have the product of 6 and
the sum or difference of -5.
Step 3: Adjust signs to get -5 & 6.
Step 4: Divide by 2 the coefficient.
Step 5: Make the signs opposite.
So, the 2 roots are x= 3/2 and x=1
1. FOIL
When two binomials are multiplied, their product can be
expanded by FOIL. The letters stand
for First, Outer, Inner, Last.
Example 1: Expand (2x + 5)(3x + 4)
Solution:
We multiply 2x by each of the terms in 3x + 4 and then
multiply 5 by each term in 3x + 4, and combine the like
terms in the middle. 2x and 3x are the First terms of each
binomial, 2x and 4 are the Outer terms, 5 and 3x are
the Inner terms, and 5 and 4 are the Last terms.
We get: (2x + 5)(3x + 4) = 6x2 + 23x + 20
Example 2: Expand (x – 3)(2x + 1)
Think of this as: (x + –3)(2x + 1)
So (x – 3)(2x + 1) = 2x2 – 5x – 3
2. Reverse FOIL" is a method used for factoring a quadratic
trinomial. The strategy is to determine the First and Last
terms of each binomial in the factored product so
the Outer and Inner products add to the middle term.
Example: Factor 3x2 + 5x – 12
Solution: First we concentrate on 3x2 and –12. 3x2 is the
product of 3x and x, so we place these First terms of the
factors:
3x2 + 5x – 12 = (3x )(x )
Now –12 could result in many different ways. It could be –
1 times 12, –2 times 6, 2 times –6, or any number of other
products.
To make it easier, we ignore the negative for the time
being and try various products that give 12, and then ask
ourselves if it is possible to get the middle term, 5x, by
subtracting the Outer and Inner products:
3x2 + 5x – 12 = (3x 2)(x 6)
Outer = 18x, Inner = 2x No
3x2 + 5x – 12 = (3x 6)(x
2)
Outer = 6x, Inner = 6x No
3x2 + 5x – 12 = (3x 4)(x 3)
Outer = 9x, Inner = 4x Yes, 9x –4x = 5x
So we want the 4x to be negative. Therefore the
factorization is:
3x2 + 5x – 12 = (3x – 4)(x + 3)
3. Factoring a GCF first
Sometimes a trinomial has a greatest common factor.
When this is the case, we should first factor that, and then
apply reverse FOIL.
Example: Factor 6x2 – 2x – 4
Solution: Every term is a multiple of 2. That is, 2 is a
greatest common factor, so we first factor that:
6x2 – 2x – 4
= 2(3x2 – x – 2)
Then we factor the trinomial in parentheses:
6x2 – 2x – 4
= 2(3x2 – x – 2)
= 2(3x + 2)(x – 1)
4. The Difference of Squares
Some quadratic binomials are the difference of two perfect
squares. When this is the case, they can be factored as a
difference times a sum, and the following formula applies:
a2 – b2 = (a – b)(a + b)
Example: Factor 9x2 – 4
Solution: 9x2 = 3x.3x and 4 = 2.2, so this binomial is the
difference of two squares. It factors as:
9x2 – 4 = (3x – 2)(3x + 2)
Example 2: Factor 3x2 – 48
Solution: 3x2 and 48 are not perfect squares, but both
have 3 as a common factor. Therefore we start by
factoring out 3:
3x2 – 48 = 3(x2 – 16)
Now x2 – 16 is the difference of squares, so it can be
factored:
3x2 – 48 = 3(x2 – 16) = 3(x – 4)(x + 4)
5. Solving Equations by Factoring
When the product of two numbers is 0, then one of the
numbers must be 0. This is a special property of 0 and
does not apply to other numbers. For example, if the
product of two numbers is 8, then neither number need be
8 (2 times 4 is 8, so the numbers could be 2 and 4).
This property of 0 is called the "zero product property"
and can be used to solve an equation when its factored
form is equal to 0.
For example, the equation (x – 5)(x + 3) = 0 tells us that the
product of the two numbers x – 5 and x + 3 is 0. Therefore
one of these numbers must be 0. That is, either x – 5 = 0 or
x + 3 = 0. The solutions to these two equations are x = 5
and x = –3, so the equation (x – 5)(x + 3) = 0 has these two
solutions.
Quadratic equations of the form ax2 + bx + c = 0 can often
be solved by factoring the left side.
Example: Solve x2 + 3 = 4x
Solution: First subtract 4x from both sides to get: x2 – 4x +
3=0
Then factor the left side: (x – 1)(x – 3) = 0
Now set each factor to 0 and solve:
x – 1 = 0 --> x = 1
x – 3 = 0 --> x = 3
Type 2 QF
Finding the Discriminant
Using the formula with D calculated before
Vedic Quadratic formula = quick w derivative
Solve x(x – 2) = 4
x(x – 2) = 4
x2 – 2x = 4
x2 – 2x – 4 = 0
• Since there are no factors of (1)(–4) = –4 that add up
to –2, then this quadratic does not factor, use the
Quadratic Formula; in this case, a = 1, b = –2, and
c = –4:
Then the answer is: x = –1.24, x = 3.24, rounded to two places.
Quadratic Formula: Just plug in the values of a,
b and c, and do the calculations. The ± means
there are TWO answers.
Discriminant
b2 - 4ac in the formula is called the Discriminant,
because it can "discriminate" between the possible types
of answer:
when b2 - 4ac is positive, we get two Real
solutions
when it is zero we get just ONE
real solution (both answers are the same)
when it is negative we get two Complex
solutions.
Using the Quadratic Formula
Put the values of a, b and c into the Quadratic Formula,
and do the calculations.
Example: Solve 5x² + 6x + 1 = 0
Coefficients are:
Quadratic Formula:
Put in a, b and c:
Solve:
a = 5, b = 6, c = 1
x = [ -b ± √(b2-4ac) ] / 2a
x = [ -6 ± √(62-4×5×1) ] / (2×5)
x = [ -6 ± √(36-20) ]/10
x = [ -6 ± √(16) ]/10
x = ( -6 ± 4 )/10
x = -0.2 or -1
Answer: x = -0.2 or x = -1
Check -0.2:
5×(-0.2)² + 6×(-0.2) + 1
= 5×(0.04) + 6×(-0.2) + 1
= 0.2 -1.2 + 1
=0
5×(-1)² + 6×(-1) + 1
= 5×(1) + 6×(-1) + 1
=5-6+1
=0
Check -1:
Complex Solutions:
When the Discriminant (the
value b2 - 4ac) is negative we get Complex solutions. It
means our answer will include Imaginary Numbers.
Example: Solve 5x² + 2x + 1 = 0
Coefficients are:
Note that The Discriminant is
negative:
Use the Quadratic Formula:
a = 5, b = 2, c = 1
b2 - 4ac = 22 - 4×5×1 = x = [ -2 ± √(-16) ] / 10
The square root of -16 is 4i (i is
So:
x = ( -2 ± 4i )/10
Answe
The graph does not cross th
complex numbers. In some
calculation, jus
Summary
Quadratic Equation in Standard Form: ax2 + bx +
c=0
Quadratic Equations can be factored
Quadratic Formula: x = [ -b ± √(b2-4ac) ] / 2a
When the Discriminant (b2-4ac) is:
positive, there are 2 real solutions
zero, there is one real solution
negative, there are 2 complex solutions