ALGEBRAIC EQUATIONS

Welcome to my Mathematics Blog.

Quadratic Equations

The equation of the fom
ax+bx+c = 0, a#0

is called a quadratic equation in a variable x. The vaues of x which satisfy
the quadratic equation are called its soiutions or roots.

Soution by Factorization
A method of finding the solutions of a quadratic equation
ax +bx +c= 0, a#0
is to find the linear factors of the quadratic expression (or function)
f(x) = ax^2+ bx +c, a #0
as fx) = glx). h(x) so that the quadratic equation becomes

g(x). h(x)= 0.

Then one or both of the factors must be equal to zero, i.e.
either g(x) = 0 or h(x) = 0.

This method of factorization gives the two solutions of the quadratic equation.
One way of factorising a quadratic expression ax + bx +c (where a,
and c are rational numbers) as a product of two linear factors with ratior
coefficients is to express a. c as a product of 2 factors obtained to break
the middle term, bx, into 2 terms. The quadratic function then becomes
an expression containing four terms which can then be factonized by the
grouping method.
This method of factorization can always be used if the expression b^2-4ac
known as the discriminant of the quadratic expression ax + be t C
perfect sq uare.

Example1

2x^2+5x=0

X(2x + 5)=0

= 0 or 2x +5 0=0

Check that these solutions satisfy the original equation.

Example2

y-9=0
(y+3) (y-3) = 0 (Difference of 2 squares)
yt3=0 or y-3 = 0
y-3 or y = 3

Check that these solutions satisfy the original equation.

Example 3

3p-P4 0

Using the general form of the coetficients
a=3, b= -1, c=-4.

Then the discriminant

b- 4ac= 49 = 7^2

We can then factorize 3p^2-p-4 as follows;

ac -12 = (-4)(3) such that -4+3=b
3p^2-4p + 3p-4=0
p(3p-4)+1(3p-4)= 0
p+1) (3p-4)=0
P+I 0 or 3p4= 0

Check that these solutions satisly the original equation

Example 4
52-5^x+1 +4=0
Put y 5^x. Then 5=(5^x)= y^2 and 5^x+1=5(5^x)=5y. Then the quadratic equation becomes
y^2-5y + 4= 0
(y-1)(y-4)=0

y- 1= 0 or y-4=0
y= I or 4
5x= 1 or 5 4
5^x=1 or xlogio 5 log10^4
0 or log10^4/1og10^5.

Check that these solutions satisfy the original equation.

Example 5

log10(X^2-5x + 94)= 2.

Rewrite the equation without log

x^2- Sx + 94= 10
x^2 5x- 6= 0
(x-6) (x+ 1)= 0

X-6 =0 or x+ 1= 0
X= 6 or-1.

Check that these solutions satisfy the original equation.

Example 6

Find the q uadratic equation in x whose roóts are 3 and4
(x-3) (x+4)=0
x^2+x-12=0.