Quadratic Equations in One Variables.pptx

Quadratic
Equation in One
Variable

Examples: 𝑥2 + 5𝑥 + 3 = 0
2𝑥2
− 4𝑥 + 9 = 0
𝑥 3 + 𝑥2 2 = 5
 𝑥 − 2 𝑥 + 3 = 0
𝑥2
− 5 = 𝑥 + 6
𝑥 𝑥 − 4 = 1
Quadratic Equation in One Variable
A quadratic equation in one variable is of the form 𝑎𝑥2 + 𝑏𝑥 + 𝑐 = 0, where
𝑎, 𝑏, and 𝑐 are real number coefficients and 𝑎 ≠ 0. This equation is in the
second degree wherein the highest exponent is 2.

𝑥 − 2 𝑥 + 3 = 0
Rewrite each equation in standard form. Then,
identify the value of 𝒂, 𝒃, and 𝒄.
1. 𝑥 − 2 𝑥 + 3 = 0
F O
I
L
F: 𝑥2
O: 3𝑥
I: −2𝑥
L: −6
Quadratic Equation:
𝑥2
− 𝑥 − 6 = 0
Therefore, 𝑎 = 1; 𝑏 = −1; 𝑐 = −6.

2. 𝑥2
− 5 = 𝑥 + 6
Quadratic Equation:
𝑥2
− 𝑥 + 1 = 0
𝑥2
− 5 = 𝑥 + 6
𝑥2
− 5 − 𝑥 = 𝑥 − 6
𝑥2
− 𝑥 − 5 + 6 = −6 + 6
𝑥2
− 𝑥 + 1
Therefore, 𝑎 = 1; 𝑏 = −1; 𝑐 = 1.

3. (2𝑥 + 1)2
= 5𝑥
Quadratic Equation:
4𝑥2
− 𝑥 + 1 = 0
(2𝑥 + 1)2
= 5𝑥
4𝑥2
+ 4𝑥 + 1 = 5𝑥
4𝑥2
+ 4𝑥 − 5𝑥 + 1 = 5𝑥 − 5𝑥
4𝑥2
− 𝑥 + 1 = 0
Therefore, 𝑎 = 4; 𝑏 = −1; 𝑐 = 1.

DRILL
1. (𝑥 + 5)2
= 3
2. 𝑥 𝑥 + 2 = 5𝑥2
− 4
3. 2𝑥 + 1 𝑥 − 3 = 𝑥 + 1
4. 𝑥 − 4 𝑥 − 6 = 2
5.
3
4
𝑥2
−
2
3
𝑥 =
3
5

The Roots of
Quadratic Equations

Nature of the Roots
If a, b, and c are real numbers and 𝐷 = 𝑏2 − 4𝑎𝑐, then the roots of the
quadratic equation 𝑎𝑥2 + 𝑏𝑥 + 𝑐 = 0 are:
I. real and unequal, if D > 0;
II. real and equal, if D = 0;
III. imaginary and unequal, if D < 0;
Moreover, the roots are:
IV. rational and unequal, if D is a perfect square;
V. irrational and unequal, if D is not a perfect square.
Discriminant (𝒃𝟐
− 𝟒𝒂𝒄)
The discriminant provided information regarding the nature of the roots of
the quadratic equation.

Determine the nature of the roots of each equation.
1. 𝑥2
− 6𝑥 + 9 = 0
𝑥2
− 6𝑥 + 9 = 0; a = 1, b = -6, and c = 9
𝐷 = 𝑏2
− 4𝑎𝑐
𝐷 = (−6)2
−4(1)(9)
𝐷 = 36 − 36
𝐷 = 0
Therefore, the roots of 𝒙𝟐
− 𝟔𝒙 + 𝟗 = 𝟎 are real, equal, and rational.

2. 2𝑥2
− 4𝑥 + 5 = 0
2𝑥2
− 4𝑥 + 5 = 0; a = 2, b = -4, and c = 5
𝐷 = 𝑏2
− 4𝑎𝑐
𝐷 = (−4)2
−4(2)(5)
𝐷 = 16 − 40
𝐷 = −24
Therefore, the roots of 𝟐𝒙𝟐
− 𝟒𝒙 + 𝟓 = 𝟎 are imaginary and unequal.

DRILL
1. 𝑥2
+ 10𝑥 + 22 = 0
2. 4𝑥2
+ 2𝑥 + 4 = 0
3. 2𝑥2
− 4𝑥 − 4 = 0
4. 𝑥2
− 10𝑥 + 22 = 0
5.
3
4
𝑥2
−
2
3
𝑥 −
3
5
= 0

Sum and Product of the Roots
Sum of The Roots = −
𝑏
𝑎
; Product of the Roots =
𝑐
𝑎
Find the sum and product of the roots of the following equations.
1. 𝑥2 − 6𝑥 + 9 = 0
𝑥2
− 6𝑥 + 9 = 0; a = 1, b = -6, and c = 9
𝑆𝑈𝑀 =
−𝑏
𝑎
𝑃𝑅𝑂𝐷𝑈𝐶𝑇 =
𝑐
𝑎
=
−(−6)
1
= 6
=
6
1
=
9
1
= 9

Find the sum and product of the roots of the following equations.
2. 2𝑥2
− 4𝑥 + 5 = 0
2𝑥2
− 4𝑥 + 5 = 0; a = 2, b = -4, and c = 5
𝑆𝑈𝑀 =
−𝑏
𝑎
𝑃𝑅𝑂𝐷𝑈𝐶𝑇 =
𝑐
𝑎
= 2
=
4
2
=
−(−4)
2
= 2.5
=
5
2

Determine the sum and products of the roots of
each equation.
DRILL
1. 𝑥2
+ 10𝑥 + 22 = 0
2. 4𝑥2
+ 2𝑥 + 4 = 0
3. 2𝑥2
− 4𝑥 − 4 = 0
4. 𝑥2
− 10𝑥 + 22 = 0
5.
3
4
𝑥2
−
2
3
𝑥 −
3
5
= 0

Solving Quadratic
Equations by:
Extracting Square Roots
Factoring
Completing The Square
Using The Quadratic Formula

No. of Solutions in a Quadratic E
quation
 Every quadratic equation in one variable has exactly two
solutions or roots.

Square Root Property
 If 𝑥2
= 𝑛, then 𝑥 = 𝑛 or 𝑥 = − 𝑛.

• If 𝑏 = 0, the equation is a pure or incomplete quadratic equation.
• Example: 𝑥2
− 4 = 0 and 3𝑥2
+ 6 = 0
• If 𝑏 ≠ 0, the equation is a complete quadratic equation.
• Example: 𝑥2
− 4𝑥 + 1 = 0 and 3𝑥2
+ 2𝑥 − 1 = 0
A. Extracting Square Roots
Step 1: Express the quadratic equation in standard form.
Step 2: Factor the quadratic expression.
Step 3: Apply the zero-product property and set each variable factor equal to 0.
Step 4: Solve the resulting linear equations.

Solve the quadratic equation by extracting the square root.
1. 𝑥2
− 11 = 0
𝑥2 − 11 = 0
𝑥2 = 11
𝑥 = ± 11
To Check: If 𝑥 = 11, If 𝑥 = − 11
( 11)2
−11 = 0 − 11
2
− 11 = 0
11 − 11 = 0 11 − 11 = 0
0 = 0 True 0 = 0 True

2. 2𝑥2
+ 3 = 39
2𝑥2
+ 3 = 39
2𝑥2
= 36
𝑥2
= 18
𝑥 = ± 18
𝑥 = ± (9)(2)
𝑥 = ± 9 2
𝑥 = ±3 2
To Check: If 𝑥 = 2 3, If 𝑥 = −3 2
2(3 2)2 + 3 = 39 −3 2
2
+ 3 = 39
2(9)(2) + 3 = 39 2 9 2 + 3 = 39
36 + 3 = 39 36 + 3 = 39
39 = 39 True 39 = 39 True

3. (𝑥 + 4)2
= 1
(𝑥 + 4)2= 1
𝑥 + 4 2 = 1
𝑥 + 4 = ±1
To Check: If 𝑥 = −3, If 𝑥 = −5
(−3 + 4)2
= 1 −5 + 4 2
= 1
(1)2= 1 (−1)2= 1
1 = 1 True 1 = 1 True
For 𝑥 + 4 = 1,
𝑥 = −3
For 𝑥 + 4 = −1,
𝑥 = −5

Solve the following quadratic equations using the
Square Root Property.
DRILL
1. 6𝑥2
− 24 = 0
2. 2𝑥2
− 34 = 0
3. (𝑥 + 2)2
= 9
4. (4𝑥 + 1)2
= 196

Zero Property of Equality
 The product 𝐴𝐵 = 0, if and only if 𝐴 = 0 or 𝐵 = 0.

B. Factoring
Step 1: Clear all fractions (if any) and write the given equation in the form
𝑎𝑥2 + 𝑏𝑥 + 𝑐 = 0.
Step 2: Factor the product of the coefficient of 𝑥 and the constant term of
the given quadratic equation.
Step 3: Express the coefficient of the middle term as the sum or difference
of the factors obtained in Step 2.
Step 4: Use zero product property to put each linear factor equal to 0.
Step 5: Solve the resulting linear equations.

Solve the quadratic equation by factoring.
1. 𝑥2
− 9𝑥 + 18 = 0
𝑥2 − 9𝑥 + 18 = 0
𝑥 − 3 𝑥 − 6 = 0
𝑥 − 3 = 0 𝑥 − 6 = 0
𝑥 = 3 𝑥 = 6
TO CHECK: 𝑥 = 3
𝑥2 − 9𝑥 + 18 = 0
32 − 9 3 + 18 = 0
9 − 27 + 18 = 0
9 − 9 = 0
0 = 0 True
TO CHECK: 𝑥 = 6
𝑥2 − 9𝑥 + 18 = 0
62 − 9 6 + 18 = 0
36 − 54 + 18 = 0
36 − 36
0 = 0 True

Solve the quadratic equation by factoring.
2. 𝑥2
+ 12 = 7𝑥
𝑥2 + 12 = 7𝑥
𝑥2 − 7𝑥 + 12 = 0
𝑥 − 3 𝑥 − 4 = 0
𝑥 − 3 = 0 𝑥 − 4 = 0
𝑥 = 3 𝑥 = 4
TO CHECK: 𝑥 = 3
𝑥2 + 12 = 7𝑥
32 + 12 = 7(3)
9 + 12 = 21
21 = 21 True
TO CHECK: 𝑥 = 4
𝑥2 + 12 = 7𝑥
42 + 12 = 7(4)
16 + 12 = 28
28 = 28 True

Solve the following quadratic equations by factoring.
DRILL
1. 𝑥2
− 5𝑥 + 6 = 0
2. 𝑥2
− 3𝑥 − 18 = 0
3. 2𝑥2
+ 3𝑥 − 2 = 0
4. 2𝑥2
+ 7𝑥 − 4 = 0

C. Completing the Square
Step 1: Rewrite the equation in the form 𝑎𝑥2
+ 𝑏𝑥 + 𝑐 = 0.
Step 2: Add to both sides the term needed to complete the square.
Step 3: Factor the perfect square trinomial.
Step 4: Solve the resulting equation by using the square root property..
• To make an expression of the form 𝑥2 + 𝑏𝑥 a perfect square
trinomial, add
𝑏
2
2
to it.

Solve the quadratic equation by completing the square.
1. 𝑥2
+ 6𝑥 + 8 = 0
𝑥2 + 6𝑥 + 8 = 0
𝑥2 + 6𝑥 = −8
𝑥2
+ 6𝑥 +
6
2
2
= −8
𝑥2 + 6𝑥 + (3)2= −8
𝑥2 + 6𝑥 + 9 = −8 + 9
𝑥 + 3 2
= 1
(𝑥 + 3)2= ± 1

1. 𝑥2
+ 6𝑥 + 8 = 0
𝑥 + 3 = ±1
𝑥 = −3 ± 1
𝑥 = −3 + 1 = −2 or 𝑥 = −3 − 1 = −4
TO CHECK: 𝑥 = −2
𝑥2 + 6𝑥 + 8 = 0
−22 + 6 −2 + 8 = 0
4 − 12 + 8 = 0
4 − 4 = 0
0 = 0 True
TO CHECK: 𝑥 = −4
𝑥2 + 6𝑥 + 8 = 0
−42 + 6 −4 + 8 = 0
16 − 24 + 8 = 0
16 − 16 = 0
0 = 0 True
SOLUTIONS: −2 & − 4

2. 𝑥2
− 2𝑥 = 1
𝑥2 − 2𝑥 = 1
𝑥2 − 2𝑥 +
2
2
2
= 1
𝑥2 − 2𝑥 + 1 2 = 1
𝑥2 − 2𝑥 + 1 = 1 + 1
(𝑥 − 1)2= 2
(𝑥 − 1)2= ± 2
𝑥 = 1 + 2 or 𝑥 = 1 − 2
SOLUTIONS:1 + 2 & 1 − 2

3. 4𝑥2
− 5𝑥 − 6 = 0
4𝑥2
− 5𝑥 − 6 = 0
4𝑥2
− 5𝑥 = 6
4 (𝑥2 −
5
4
𝑥) = 6
4 (𝑥2 −
5
4
𝑥)
4
=
6
4
𝑥2 −
5
4
𝑥 =
3
2

3. 4𝑥2
− 5𝑥 − 6 = 0
Complete the Square:
5
4
2
2
=
5
8
2
=
25
64
𝑥2
−
5
4
𝑥 +
25
64
=
3
2
+
25
64
Add the fractions on the right side.
3
2
+
25
64
=
96
64
+
25
64
=
121
64

3. 4𝑥2
− 5𝑥 − 6 = 0
𝑥2 −
5
4
𝑥 +
25
64
=
121
64
𝑥 −
5
8
2
=
121
64
𝑥 −
5
8
2
= ±
121
64

3. 4𝑥2
− 5𝑥 − 6 = 0
𝑥 −
5
8
= ±
11
8
𝑥 =
5
8
±
11
8
𝑥 =
5
8
+
11
8
=
16
8
= 2
𝑥 =
5
8
−
11
8
= −
6
8
= −
3
4 SOLUTIONS: 2 & −
3
4

Solve the following quadratic equations by
completing the square.
DRILL
1. 𝑥2
− 2𝑥 − 8 = 0
2. 𝑥2
+ 7𝑥 − 18 = 0
3. 3𝑥2
= 2 − 𝑥
4. 𝑥2
= 9

D. Using the Quadratic Formula
𝑥 =
−𝑏 ± 𝑏2 − 4𝑎𝑐
2𝑎
𝑥 =
−𝑏 ± 𝑏2 − 4𝑎𝑐
2𝑎
𝑥 =
−9 ± 92 − 4(1)(14)
2(1)
Solve the quadratic equation using the quadratic formula.
1. 𝑥2
+ 9𝑥 + 14 = 0

1. 𝑥2
+ 9𝑥 + 14 = 0
𝑥 =
−9 ± 81 − 56
2
𝑥 =
−9 ± 25
2
𝑥 =
−9 ± 5
2

1. 𝑥2
+ 9𝑥 + 14 = 0
𝑥 =
−9 ± 5
2
𝑥 =
−9+5
2
or 𝑥 =
−9−5
2
𝑥 =
−4
2
or 𝑥 =
−14
2
𝑥 = −2 or 𝑥 = −7

Solve the following quadratic equations using the
quadratic formula.
DRILL
1. 𝑥2
− 2𝑥 + 1 = 0
2. 𝑥2
− 6𝑥 + 9 = 0
3. 𝑥2
− 5𝑥 − 5 = 0
4. 3𝑥2
− 5𝑥 − 2 = 0

Equations in Quadratic
Form (𝑎𝑓2
+ 𝑏𝑓 + 𝑐 = 0)

Equations in Quadratic Form
An equation in quadratic form is any equation of the form 𝑎𝑓2 + 𝑏𝑓 + 𝑐 = 0,
where 𝑓 is any algebraic expression. These equations can be solved using
simple substitution.
Example: Solve the following equations.
1. 9𝑥4 − 52𝑥2 + 64 = 0
In 9(𝑥2
)2
−52 𝑥2
+ 64 = 0, let 𝑓 = 𝑥2
.
9𝑓2 − 52𝑓 + 64 = 0
𝑓 − 4 9𝑓 − 16 = 0
𝑓 − 4 = 0 or 9𝑓 − 16 = 0
𝑓 = 4 or 𝑓 =
16
9

1. 9𝑥4
− 52𝑥2
+ 64 = 0
Let us solve for 𝑥.
𝑥2
= 𝑓 = 4 or 𝑥2
= 𝑓 =
16
9
𝑥2 = 4 or 𝑥2 =
16
9
𝑥 = ± 4 or 𝑥 = ±
16
9
𝑥 = ±2 or 𝑥 = ±
4
3
Solution Set: −2, −
4
3
,
4
3
, 2

2. (2𝑥 − 1)2
− 6 2𝑥 − 1 = −8
In(2𝑥 − 1)2−6 2𝑥 − 1 = −8, let 𝑔 = 2𝑥 − 1.
𝑔2
− 6𝑔 = −8
𝑔2
− 6𝑔 + 8 = 0
𝑔 − 4 𝑔 − 2 = 0
𝑔 − 4 = 0 or 𝑔 − 2 = 0
𝑔 = 4 or 𝑔 = 2

2. (2𝑥 − 1)2
− 6 2𝑥 − 1 = −8
Let us solve for 𝑥.
2𝑥 − 1 = 𝑔 = 4 or 2𝑥 − 1 = 𝑔 = 2
2𝑥 − 1 = 4 or 2𝑥 − 1 = 2
2𝑥 = 5 or 2𝑥 = 3
𝑥 =
5
2
or 𝑥 =
3
2
Solution Set:
5
2
,
3
2

Solve for 𝒙 in each equation.
DRILL
1. 𝑥4
+ 4 = 5𝑥2
2. 𝑥4
− 17𝑥2
+ 16 = 0

Rational Algebraic
Equations Transformable
to Quadratic Equation

Rational Algebraic Equations
Rational algebraic equations are equations that contain one or more
rational algebraic expressions (algebraic expressions in the form
𝑛
𝑑
,
where 𝑛 and 𝑑 are polynomials and 𝑑 ≠ 0). These equations can be
reduced to polynomial equations by clearing off fractions.
Examples:
5𝑥
6
−
1
2
=
1
3
𝑥
2
+ 3𝑥 = 7
4𝑥−6
2𝑥−3
=
7
𝑥+1

But, take note, not all rational algebraic
equations can be transformed into quadratic
equations.
Rational Algebraic Equations
Transformable to Quadratic Equations
1. It has to be a rational equation.
2. There should only be one variable.
3. If the expressions are linear, then there should be a variable on both
numerator and denominator. Example:
4𝑥−6
2𝑥−3
4. If the denominators are all constant, then the numerator should have at
least one variable that has a power of 2. Example:
𝑥2+1
7

Solving Rational Algebraic Equations
Transformable to Quadratic Equations by
Finding the LCD
1. Multiply both sides of the equation by the LCD.
2. Apply the distributive law.
3. Combine like terms.
4. Apply the inverse operation.
5. Rewrite the equation in standard form.
6. Solve for the resulting quadratic equation using any of the 4 methods.

1.
6
𝑥
+
𝑥−3
4
= 2
24 + 𝑥 𝑥 − 3 = 8𝑥
24 + 𝑥2
− 3𝑥 = 8𝑥
24 + 𝑥2
− 3𝑥 − 8𝑥 = 0
24 + 𝑥2
− 11𝑥 = 0
𝑥2
− 11𝑥 + 24 = 0
6
𝑥
+
𝑥−3
4
= 2
[4𝑥] [4𝑥]
24𝑥
𝑥
+
4𝑥(𝑥−3)
4
= 8𝑥

1.
6
𝑥
+
𝑥−3
4
= 2  𝑥2
− 11𝑥 + 24 = 0
𝑥2
− 11𝑥 + 24 = 0
𝑥 − 3 𝑥 − 8 = 0
𝑥 − 3 = 0 or 𝑥 − 8 = 0
𝑥 = 3 or 𝑥 = 8

Solving Rational Algebraic Equations
Transformable to Quadratic Equations
using Cross-Multiplication
1. Cross-multiply the means and the extremes.
2. Apply the inverse operation.
3. Rewrite the equation in standard form.
4. Solve for the resulting quadratic equation using any of the 4 methods.
Cross-Multiplication Property:
If
𝑎
𝑏
=
𝑐
𝑑
, then 𝑎𝑑 = 𝑏𝑐.
then 𝑎𝑑 = 𝑏𝑐.

1.
4𝑥−6
2𝑥−3
=
7
𝑥+1
4𝑥2
− 2𝑥 − 6 = 14𝑥 − 21
4𝑥2
− 2𝑥 − 14𝑥 − 6 + 21
= 14𝑥 − 14𝑥 − 21 + 21
4𝑥2
− 16𝑥 + 15 = 0
4𝑥2
− 6𝑥 − 10𝑥 + 15 = 0
4𝑥 − 6
2𝑥 − 3
=
7
𝑥 + 1
4𝑥 − 6 𝑥 + 1 = 7(2𝑥 − 3)
EXTREMES MEANS
EXTREMES
MEANS

1.
4𝑥−6
2𝑥−3
=
7
𝑥+1
 4𝑥2
− 16𝑥 + 15 = 0
4𝑥2
− 16𝑥 + 15 = 0
(4𝑥2
− 6𝑥)(−10𝑥 + 15) = 0
2𝑥 2𝑥 − 3 − 5 2𝑥 − 3 = 0
2𝑥 − 3 2𝑥 − 5 = 0
2𝑥 − 3 = 0 or 2𝑥 − 5 = 0
2𝑥 = 3 or 2𝑥 = 5
𝑥 =
3
2
or 𝑥 =
5
2

Solve the following equations.
DRILL
1.
10
𝑥
+ 1 = 3𝑥

Problem Solving
involving
Quadratic Equations

PROBLEM # 1: The sum of two numbers is 22 and the sum of their
squares is 250. Find the numbers.
STEP 1: Define the variables.
Let 𝑥 = one of the two numbers
22 − 𝑥 = the other number
STEP 2: Identify the thought process.
one of the
two numbers
2
+ the other
number
2
= 250

PROBLEM # 1: The sum of two numbers is 22 and the
sum of their squares is 250. Find the numbers.
STEP 3: Formulate the equation.
𝑥2
+ (22 − 𝑥)2
= 250
STEP 4: Solve the equation.
𝑥2
+ (22 − 𝑥)2
= 250
𝑥2
+ 484 − 44𝑥 + 𝑥2
= 250
2𝑥2
− 44𝑥 + 234 = 0
2𝑥2
− 44𝑥 + 234
2
=
0
2

PROBLEM # 1: The sum of two numbers is 22 and the
sum of their squares is 250. Find the numbers.
𝑥2
− 22𝑥 + 117 = 0
(𝑥 − 13)(𝑥 − 9) = 0
𝑥 = 13 𝑜𝑟 𝑥 = 9
𝐼𝑓 𝑥 = 13, 𝑡ℎ𝑒𝑛 22 − 13 = 9.
𝐼𝑓 𝑥 = 9, 𝑡ℎ𝑒𝑛 22 − 9 = 13.
The numbers are 9 and 13.
To Check: 9 + 13 = 22
92
+ 132
= 81 + 169 = 250

Solve the word problem below.
DRILL
PROBLEM: Two numbers differ by 9. The sum of their squares is 653.
Identify the 2 numbers.
Let 𝑥 = the smaller number
𝑥 + 9 = the larger number

PROBLEM # 2: If the length of each side of a square is increased by 5
cm, the area is multiplied by 4. What is the length of the original
side of the square?
STEP 1: Define the variables.
Let 𝑠 = one of the two numbers
𝑠2
= the other number
STEP 2: Identify the thought process.
one of the
two numbers + 5
2
= 4
area of
the square

PROBLEM # 2: If the length of each side of a square is
increased by 5 cm, the area is multiplied by 4. What is
the length of the original side of the square?
(𝑠 + 5)2
= 4(𝑠2
)
(𝑠 + 5)2
= 4(𝑠2
)
𝑠2
+ 10𝑠 + 25 = 4𝑠2
−4𝑠2
+ 𝑠2
+ 10𝑠 + 25 = 4𝑠2
− 4𝑠2
0 = 3𝑠2
− 10𝑠 − 25
0 = 3𝑠2
+ 5𝑠 − 15𝑠 − 25
0 = (3𝑠2
+ 5𝑠)(−15𝑠 − 25)

PROBLEM # 2: If the length of each side of a square is
increased by 5 cm, the area is multiplied by 4. What is
the length of the original side of the square?
0 = 3𝑠2
+ 5𝑠 −15𝑠 − 25
0 = 𝑠 3𝑠 + 5 − 5 3𝑠 + 5
0 = (3𝑠 + 5)(𝑠 − 5)
3𝑠 + 5 = 0 𝑜𝑟 𝑠 − 5 = 0
𝑠 = −
5
3
𝑜𝑟 𝑠 = 5
Since a side of the square is a physical quantity that cannot assume a
negative value, we can only accept 𝑠 = 5. Therefore, the length of the
original side of the square is 5 cm.

PROBLEM # 3: A rectangle is 30 cm long and 20 cm wide. A rectangular
strip added to one side and another of the same width to the other side
result in the doubling of the area. Find the width of the strip.
STEP 1: Draw the figure.

strip added to one side and another of the same width to the other
side result in the doubling of the area. Find the width of the strip.
30 + 𝑤 20 + 𝑤 = 2(30)(20)
600 + 50𝑤 + 𝑤2
= 2(30)(20)
𝑤2
+ 50𝑤 + 600 = 1200
𝑤2
+ 50𝑤 + 600 − 1200 = 1200 − 1200
Let 𝒘 = width of the strip
STEP 2: Define the variable.

𝑤2
+ 50𝑤 + 600 − 1200 = 0
𝑤2
+ 50𝑤 − 600 = 0
𝑤 + 60 𝑤 − 10 = 0
𝑤 + 60 = 0 𝑜𝑟 𝑤 − 10 = 0
𝑤 = −60 𝑜𝑟 𝑤 = 10
Since the width cannot be negative, we can only accept 𝑤 = 10. Therefore,
the width of the rectangular strip is 10 cm.
strip added to one side and another of the same width to the other
side result in the doubling of the area. Find the width of the strip.

Solve the word problem below.
DRILL
PROBLEM: The perimeter of a rectangle is 60 m and its area is
𝟐𝟐𝟏 𝒎𝟐
. Find the length and the width of the rectangle.

Quadratic Equations in One Variables.pptx

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