# Quadratic Equations

**Definition:** A quadratic equation is an equation
involving an

unknown to the second power (degree), but no higher power. The

general form of the quadratic equation is

ax ^{2} + bx + c = 0, where x is the unknown and a ≠ 0.

The coefficient of the unknown of the highest exponent is
called the

**leading coefficient.** For example, the leading coefficient of the

expression 7x^{2} - 5x + 4 is 7, while the leading coefficient of the

expression 10+ 3u- 5u^{2} is - 5.

Our first inclination is to think of quadratic as dealing
with "four".

But the dictionary's definition of quadratic is "containing quantities of

the second degree". The name comes from "quadrature", an ancient

problem of constructing a square having the same area as the given

figure using only a compass and a straight edge. (To see that

quadrature is not such a simple problem , try constructing a square

having the same area as the rectangle whose length is 5 inches and

whose width is 2 inches, using only a compass and a straight edge.)

**7.1 Solving Quadratic Equations by Factoring
**To solve quadratic equations by factoring:

a) write the equation in standard form (bring all terms to the

left -hand side so that the right-hand side is zero),

b) factor the expression on the left (or right) side,

c) solve for the values of the unknown using the multiplication

property of zero .

If A B = 0, then either A = 0 or B = 0. There are no other

possibilities.

But if A B = 1 (or any other non- zero number ), there are infinite

possibilities. For example,

A = 1, B = 1

etc.

__ Example 1:__ Find the solutions of the
equation 2x

^{2}= 5x + 3 .

solution: 2x

^{2}- 5x - 3 = 0 bring all terms to the left-hand

side so that the

right-hand side is zero

(2x + 1)(x - 3) = 0 factor the trinomial

2x + 1 = 0 or x - 3 = 0 multiplication property of

zero

2x = - 1

So, and 3 are the solutions.[Note: Equation of degree 2

has , in general, 2 solutions.]

**Example 2: **Find the solutions of the equation 12n^{2}
= 9n .

solution: 12n^{2} - 9n = 0 write equation in standard

form

3n(4n - 3) = 0 factor out common factors

3n = 0 or 4n - 3 = 0 set each factor equal to zero

The solutions are 0 and .

**Example 3: **Find the solutions: y^{2} - 16 =
0

solution: y^{2} - 16 = 0 the right-hand side is already

equal to zero

(y + 4)(y - 4) = 0 factor the difference of

squares

y + 4 = 0 or y - 4 = 0

y = - 4 or y = 4

The solutions are -4 and 4, which can also be written as .

**Exercises 7.1**

Solve the following quadratic equations by factoring:

(a) x^{2} - 7x + 6 = 0

(b) n^{2} + n = 6

(c) y^{2} + 2y = 0

(d) 3m = m^{2}

(e) 9t^{2} - 1 = 0

(f) 4g^{2} = 25

(g) 2a^{2} = 5a - 3

(h) 3 + w = 2w^{2}

(i) 12x^{2} = 27

(j) 12x^{2} = 27x

(k) 2n^{2} + 10 = 12n

(l) 6y^{2} = 6 - 5y

(m) (a - 3)(2a + 1) = 39

(n) 3p(2p - 1) = 18

(o) (4m - 3)(m - 2) = 6

(p) (2t + 5)(8t - 5) = 30t

(q) 6g^{2} + 56 = 37g

(r) 1 = 64K^{2}

(u) 0.09n^{2} = 36

(v) y^{2} - 20 = y

**7.2 Solving Quadratic Equations by Extracting Roots
**Recall the earlier section on the Pythagorean Theorem: a

^{2}+ b

^{2}=

c

^{2}. If a = 3 and b = 4, then

We took the square root (extracted the root) of 25 to get
5. But is 5

the only answer? If , then c = 5 or c = - 5 (since after squaring a

positive or a negative number, we always get a positive value). For

the problem above we were interested in the of
side c,

therefore, the positive value is automatically chosen. In this section,

we find ALL possible solutions--both positive and negative.

Equations of a more general forms x^{2} - c = 0 or
(ax - n)^{2} - c = 0,

where c is a positive number, can be solved by root extraction.

Actually this method is an extension of the method covered in the

preceding section .

We have been saying that the expression is not factorable.
The

statement is not entirely correct. We should say that is not

factorable **using only integers.** If we allow numbers like square

roots of positive numbers to be used , then the expression becomes

factorable because 3 can be written as , and so we can factor as

=

=

So, if we had the problem of finding all solutions of the
equation , we

bring all terms to the left-hand side and factor the left-hand side, as in

the preceding section:

= 0

The solutions are and , which can be written as .

It is easy to see that we can replace 3 with any positive
number. So,

we have the following result:

For any positive number a, if , then .

**Example 1: **Find the solutions: y^{2} - 16 =
0

solution: isolate the squared variable

The solutions are .

**Example 2:** Solve the equation: 12n^{2} - 9
= 0

solution: 12n^{2} = 9 isolate squared term

The solutions are

and

, which can also be written as and.

**Example 3 :** Solve the equation: 3(x - 1)^{2}
= 24

solution: 3(x - 1)^{2} = 24 squared quantity is isolated

(x - 1)^{2} = 8

The solutions are and , which can also be written as

and .

"Why can't we use extraction of roots for other forms of
the quadratic

equation?"

ax^{2} + bx + c = 0: 2x^{2} + 3x + 1 = 0

2x^{2} = - 3x - 1

ax^{2} + bx = 0: x^{2} - 5x = 0

x^{2} = 5x

**Exercises 7.2
**Solve the following quadratic equations by extracting roots:

(a) x

^{2}- 4 = 0

(b) n

^{2}= 25

(c) p

^{2}- 2 = 0

(d) m

^{2}= 7

(e) a

^{2}- 12 = 0

(f) K

^{2}= 27

(g) 2y

^{2}- 98 = 0

(h) 3z

^{2}= 48

(i) 3w

^{2}= 30

(j) 2t

^{2}- 126 = 0

(k) 5x

^{2}= 1

(l) 2a

^{2}- 7 = 0

(m) 4p

^{2}= 1

(n) 25m

^{2}= 9

(o) 12k

^{2}- 5 = 0

(p) 24z

^{2}= 8

(u) 2(7w + 3)

^{2}= 128

(v) 3(2y - 5)

^{2}= 24

**7.3 Solving Quadratic Equations by the Quadratic**

Formula

Formula

The

*quadratic formula*can be derived by solving the general

quadratic equation

ax

^{2}+ bx + c = 0 using the method known as completing the square

(which is not discussed here).

When a procedure is often repeated, mathematicians prefer
to use a

formula to shorten the number of steps . Use of the quadratic formula

produces results more quickly than by the method of completing the

square. In using the formula, you want to:

a) write the equation in standard form,

b) correctly identify a, b, and c,

c) carefully substitute into the formula .

**Example 1 :** Find the solutions of the equation x^{2}
+ 4x - 6 = 0

Solution: x^{2} + 4x - 6 = 0 equation is in standard form;

hence, a = 1, b = 4, c = - 6

Therefore, .

**Example 2:** Find the solutions of the equation: 2n^{2}
= n + 3

Solution: 2n^{2} - n - 3 = 0 write the equation in

standard form and substitute

into the formula

[Note: When the solutions are rational (as in this case), the equation

could have been solved by factoring. Try this and see if you obtain

the same results.]

__ Example 3:__ Solve the equation: 2 + 2w + w

^{2}= 0

Solution: w

^{2}+2w + 2 = 0 write the equation in standard

form and

substitute into the formula

But, since the square root of - 4 is not a real number

we say that **there are no real solutions.**

In the previous examples, we used the formula with
equations whose

left-hand side contained a trinomial expression. What if the left-hand

side is a quadratic binomial ?

__ Example 4:__ Solve the equation: 8x

^{2}+ 4x = 0

Solution: If we wish to use the quadratic formula, then a = 8, b = 4,

and c = 0.

But since the left-hand expression is easily factorable,

there is no need to use the formula.

4x(2x + 1) = 0

4x = 0 or 2x + 1 = 0

__ Example 5 : __Solve the equation: 12y

^{2}- 9 = 0

Solution: If we wish to use the quadratic formula, then a = 12, b =

0, and c = - 9.

But since we can use the extraction of roots method, there

is no need to use the formula.

Although the quadratic formula can be used to solve all
quadratic

equations, it is often the most time-consuming. We usually reserve

the use of the quadratic formula for those situations when factoring

or extracting roots do not work.

**Exercises 7.3
**Solve the following quadratic equations by the most effective method.

(a) x

^{2}- 8x + 15 = 0

(b) 2n

^{2}+ 3n = 2

(c) 6y

^{2}- 14y = 0

(d) 6z

^{2}= 48

(e) 2a

^{2}= 3a + 1

(f) m

^{2}+ 10m = 4

(i) n

^{2}+ 8n = 1

(j) 2z

^{2}= 3z - 8

(k) 4a

^{2}+ 5a = 0

(l) 3m

^{2}+ 4m = 3

(m) 4t

^{2}= 4t + 7

(n) 12w

^{2}- 27 = 0

(o) x

^{2}- 2x = 5 + 2x

(p) 2y

^{2}+ 2y - 3 = 5 - y

^{2}

(q) (n + 1)

^{2}= 2 - 4n

(r) (2x - 1)

^{2}= 12

(s) (2x + 1)

^{2}= 2 + 8x

(t) 2(w - 1)

^{2}= 7w

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