Solving equations of degree 2 : QUADRATIC



The general form of a quadratic equation is $ax^2+bx+c=0$

The equation has no solution if $b^2<4ac$

The equation has exactly one solution if $b^2=4ac$

This equation has 2 solutions given by $\frac{-b \pm \sqrt{b^2-4ac}}{2a}$ if $b^2>4ac$

The sum of roots is $\frac{-b}{a}$

The product of roots is $\frac{c}{a}$

If the roots are $r_1$ and $r_2$, the equation can be written as $(x-r_1)(x-r_2)=0$

A quick way to solve a quadratic, without the above formula is to factorize it :
Step 1> Divide throughout by coeff of x^2 to put it in the form $x^2+dx+e=0$
Step 2> Sum of roots = -d and Product = e. Search for 2 numbers which satisfy this criteria, let them be f,g
Step 3> The equation may be re-written as (x-f)(x-g)=0. And the solutions are f,g

Eg. $x^2+11x+30=0$
The sum is -11 and the product is 30. So numbers are -5,-6
$x^2+11x+30=x^2+5x+6x+30=x(x+5)+6(x+5)=(x+5)(x+6)$

Solving equations with DEGREE>2

You will never be asked to solved higher degree equations, except in some cases where using simple tricks these equations can either be factorized or be reduced to a lower degree or both. What you need to note is that an equation of degree n has at most n unique solutions.

Factorization
This is the easiest approach to solving higher degree equations. Though there is no general rule to do this, generally a knowledge of algebraic identities helps. The basic idea is that if you can write an equation in the form :

$A*B*C=0$

where each of A,B,C are algebraic expressions. Once this is done, the solution is obtained by equating each of A,B,C to 0 one by one.

Eg. $x^3 + 11x^2 + 30x = 0$
$x * (x^2 + 11x + 30) = 0$
$x * (x+5) * (x+6) = 0$

So the solution is x=0,-5,-6

Reducing to lower degree

This is useful sometimes when it is easy to see that a simple variable substitution can reduce the degree.

Eg. $x^6 -3x^3 + 2 = 0$
Here let $y=x^3$
$y^2-3y+2=0$
$(y-2)(y-1)=0$

So the solution is y=1 or 2 or x^3=1 or 2 or x=1 or $\sqrt[3]{2}$

Other tricks

Sometimes we are given conditions such as the variables being integers which make the solutions much easier to find. When we know that the solutions are integral, often times solutions are easy to find using just brute force.

Eg. $a^2+b^2=116$ and we know a,b are integers such that a<b
We can solve this by testing values of a and checking if we can find b
a=1 b=root(115) Not integer
a=2 b=root(112) Not integer
a=3 b=root(107) Not integer
a=4 b=root(100)=10
a=5 b=root(91) Not integer
a=6 b=root(80) Not integer
a=7 b=root(67) Not integer
a=8 b=root(52)<a
So the answer is (4,10)

Algebraic Identities

These can be very useful in simplifying & solving a lot of questions :

• $(x+y)^2=x^2+y^2+2xy$
• $(x-y)^2=x^2+y^2-2xy$
• $x^2-y^2=(x+y)(x-y)$
• $(x+y)^2-(x-y)^2=4xy$
• $(x+y+z)^2=x^2+y^2+z^2+2(xy+yz+zx)$
• $x^3+y^3=(x+y)(x^2+y^2-xy)$
• $x^3-y^3=(x-y)(x^2+y^2+xy)$