## Solving Quadratic Equations by Factoring

In the previous post, we have learned how to solve quadratic equations by extracting the roots. In this post, we are going to learn how to solve quadratic equations by factoring.

To solve quadratic equations by factoring, we need to use the zero property of real numbers. It states that the product of two real numbers is zero if at least one of the two real numbers is zero. In effect, we need to transfer all the terms to the lefth hand side, let the right hand side be 0, equate factored form zero and find the value of x.

Example 1: $2x(x + 5) = 0$

Solution
$2x = 0, x = 0$
$x + 5 = 0, x = -5$.

This means the solutions of $2x(x+5) =0$ are $0$ or $-5$.

Example 2: $x^2 - x = 6$

Solution
Subtracting $6$ from both sides, we have
$x^2 - x - 6 = 0$.

Factoring, we have
$(x - 3)(x + 2) = 0$
$x - 3 = 0, x = 3$
$x + 2 = 0, x = -2$.

This means the solutions of $x^2 - x = 6$ are $3$ or $-2$.

Example 3: $x^2 = x \sqrt{3}$

Solution

Subtracting $\sqrt{3}$ from both sides,
$x^2 - x \sqrt{3} = 0$.

Factoring out $x$, we have
$x(x - \sqrt{3}) = 0$.

Equating both expression to 0, we have
$x = 0$
$x - \sqrt{3} = 0, x = \pm \sqrt{3}$.

This means the solutions of $x^2 - x = 6$ are $0$ or $\pm \sqrt{3}$.

Example 4: Solve $x^2 = 16$.

Solution

$x^2 - 16 = 0$
$(x + 4)(x - 4) = 0$
$x = -4, x = 4$.

This means the solutions of $x^2 = 16$ are $-4$ or $4$.

Example 5: Solve $x^2 + 2x + 1 = 0$.

Solution
$x^2 + 2x + 1 = 0$
$(x + 1)(x + 1) = 0$
$x + 1 = 0, x = -1$
$x + 1 = 0, x = -1$
$x = -1, x = -1$.

This means the solutions of $x^2 + 2x + 1 = 0$ are $katex -1$ or $-1$.

Example 6: Solve $(x - 3)^2 = 4x$

Solution
$(x - 3)^2 = 4x$
$x^2 - 6x + 9 = 4x$
$x^2 - 6x - 4x + 9 = 0$
$x^2 - 10x + 9 = 0$
$(x - 1)(x - 9) = 0$
$x = 1, x = 9$.

This means the solutions of $(x - 3)^2 = 4x$ are $1$ or $9$.

Example 7: $\frac{2x - 3}{x + 3} = \frac{x + 3}{x - 3}$
Solution

Cross multiplying, we have
$(2x - 3)(x - 3) = (x + 3)(x + 3)$.

Expanding, we have
$2x^2 - 9x + 9 = x^2 + 6x + 9$.

Transposing all the terms to the left hand side, we have
$x^2 - 15x = 0$
$x(x - 15) = 0$
$x = 0$
$x - 15 = 0$
$x = 15$.

This means the solutions are $0$ or $15$.

## Solving Quadratic Equations by Extracting the Square Root

In the previous post, we have learned about quadratic equations or equations of the form $ax^2 + bx + c = 0$, where a is not equal to 0. In this equation, we want to find the value of x which we call the root or the solution to the equation.

There are three strategies in finding the root of the equation: by extracting the roots, by completing the square, and by the quadratic formula. In this example, we will discuss, how to find the root of the quadratic equation by extracting the root.

Just like in solving equations, if we want to find the value of x, we put all the numbers on one side, and all the x’s on one side. Since quadratic equations contain the term $x^2$, we can find the value of x by extracting the square roots. Below are five examples on how to do this.

Example 1: $2x^2 = 0$

Solution

Dividing both sides by 2, we have

$\frac{2x^2}{2} = \frac{0}{2}$.

This gives us

$x^2 = 0$.

Extracting the square root of both sides, we have

$x = 0$.

Therefore, the root $x = 0$.

Example 2: $x^2 - 36 = 0$

Solution

$x^2 - 36 = 0$.

Adding 36 to both side, we have

$x^2 - 36 + 36 = 0 + 36$.

$x^2 = 36$.

Extracting the square root of both sides, we have

$\sqrt{x^2} = \sqrt{36}$.

$x = \pm 6$.

In this example, x has two roots: x = 6 and x = -6.

Example 3: $x^2 + 81 = 0$

Solution

Subtracting 81 from both sides, we have

$x^2 + 81 - 81 = 0 - 81$

$x^2 = -81$

$\sqrt{x^2} = - 81$

$x = \sqrt{-81}$.

In this case, there is no number that when multiplied by itself is negative. For example, negative times negative is equal to positive, and positive times positive is equal to positive. Therefore, there is no real root. There is, however, what we call a complex root as shown in the video below.

Example 4: $5x^2 = 12$

Solution

$5x^2 = 12$

Dividing both sides by 5, we have

$x^2 = \frac{12}{5}$.

Extracting the square root of both sides, we have

$\sqrt{x^2} = \sqrt{\frac{12}{5}} = \frac{\sqrt{12}}{\sqrt{5}}$

$x = \frac{\sqrt{4}\sqrt{3}}{\sqrt{5}}$

$x = \frac{2 \sqrt{3}}{\sqrt{5}}$.

Example 5: $3x^2 = - 4$

$3x^2 = -4$

Dividing both sides by 3, we have

$x^2 = \frac{-4}{3}$.

Extracting the square root of both sides, we have

$\sqrt{x^2} = \sqrt{\frac{-4}{3}}$.

Again, the sign of the number inside the radical is negative, so there is no real root. To know how to compute for the complex root, watch the video below.

The length of a rectangle is 3 cm more than its width. Its area is equal to 54 square centimeters. What is its length and width?

Solution

Let

x = width of rectangle
x + 3 = length of rectangle

The area of a rectangle is the product of the length and width, so we have

Area= x(x + 3)

which is equal to 54.

Therefore, we can form the following equation:

x(x + 3) = 54.

By the distributive property, we have

$x^2 + 3x = 54$

Finding the value of x

In the equation, we want to find the value of x that makes the equation true. Without algebraic manipulation, we can find the value of x by assigning various values to x. The equation $x^2 + 3x = 54$ indicates that one number is greater than the other by 3 and their product is 54. Examining the numbers with product as 54, we have,

1 and 54
2 and 27
3 and 18
6 and 9.

Note: We have excluded the negative (e.g. (-1)(-54) = 54) numbers since a side length cannot be negative.

Now, 9-6 = 3 which means that the side lengths of the rectangle are 6 and 9. Yes, their product is 54 and one is 3 greater than the other.

In the equation above, subtracting both sides by 54, we have

$x^2 + 3x - 54 = 54 - 54$

$x^2 + 3x - 54 = 0$.

The equation that we formed above is an example of a quadratic equation.

A quadratic equation is of the form $ax^2 + bx + c = 0$, where a, b, and c are real numbers and a not equal to 0. In the example above, a = 1, b = 3, and c = -54.

In the problem above, we got the value of x by testing several values, however, there are more systematic methods. In the next post, we will be discussing one of these methods. These methods are factoring, completing the square, and quadratic formula.

## FREE Civil Service Examination Reviewers

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Again, good luck to all examinees.

## October 2015 Civil Service Exam List of Passers

Below are links to the list of passers in the October 2015 Paper and Pencil Test on Subprofessional and Professional Civil Service Examination. Congratulations to all passers.

Professional Exam

Region 3
Region 4 (A-D), (E-M), (N-Z)
Region 5
Region 6 (A – K) (L – Z)
Region 7
Region 8
Region 9
Region 10
Region 11
Region 12
CARAGA
ARMM

Subprofessional Exam

Note: Although the lists were directly lifted from the CSC website due to traffic congestion, you may verify your name in the original list.

## October 2015 Civil Service Subprof Exam Result – ARMM List of Passers

Below is the list of passers from ARMM of the Civil Service Exam Subprofessional Examination conducted on October 18, 2015.

Seq. No. ExamNo Name
1 940349 ALI, REHAN A
2 940345 BOFILL, SHALEE MAE P
3 940303 BUSTAMANTE, ELIZABETH E
4 940320 PANDA, CHARLIMAGNE G
5 940301 BOFILL, KIMBERLY ANN P
*** NOTHING FOLLOWS ***

Congratulations!

## October 2015 Civil Service Prof Exam Result – ARMM List of Passers

Below is the list of passers from ARMM of the Civil Service Exam Professional Examination conducted on October 18, 2015.

Seq No. Region ExamNo Name
1 16 535186 ABATAYO, SHARIFFA AINIE M
2 16 568345 ABBAS, FATIMAH FARMAILAH P
3 16 568558 ABDULRAHIM, NIHAYA D
4 16 568721 ABSIN, GLEZY P
5 16 568768 ACMALI, AMINA D
6 16 569074 ADIONG, SHARIFA MUJANNAH B
7 16 535179 AKMAD, FAYRUZ HYNE S
8 16 568938 ALAUYA, AZIZAH L
9 16 535307 ALI, ERSHAD B
10 16 535138 ALONTO, AMEEN ANDREW L
11 16 568398 AMPA, MO-AMAR S
12 16 568550 AMPA, MORSIDAH S
13 16 568417 ANGIN, AKIMAH HU
14 16 569082 ATING, MOHAMMAD HAFFIZ S
15 16 568966 BADRON, JEHAN A
16 16 568336 BANDRANG, NOR-HANN A
17 16 568849 BARODI, AL-AYSHA S
18 16 568786 BARRAT, SALEM M
19 16 568263 CALAUTO, SHAHANIFA L
20 16 568576 CASAN, MOHAMMAD KHALID R
21 16 535633 CASTILLON, ANALIE A
22 16 535640 CODILLA, SOARRAINE O
23 16 569002 DAHONOG, EILEEN JOYCE G
24 16 568231 DATAMAN, JUBAIRAH S
25 16 568476 DIANALAN, AS-HARRIE D
26 16 568621 DIMAL, ASHRAF G
27 16 568615 DIMAL, JERONISSA G
28 16 568270 DIMAPORO, JOHAINA B
29 16 568362 DIRANGGARUN, NORHANA M
30 16 568684 DITUCALAN, NASSIF M
31 16 568759 DOMAOT, ASRIFAH P
32 16 568714 DONAIRE, KEVIN S
33 16 568813 FACUNDO, JUN IAN A
34 16 568840 GANI, SITIE ASHIA D
35 16 535249 GURO, MOHAMAD ANUAR G
36 16 568722 HADJI SOCOR, ABDUL MOJIB M
37 16 568700 HADJI-YUSOP, NORHANAN A
38 16 535326 HARON, SAMRUDIN Z
39 16 568306 IBRAHIM, ISA L
40 16 568875 IBRAHIM, NORODIN H.I
41 16 535187 ILISAN, MA ODESA ANN S
42 16 535441 LAO, AMJAD A
43 16 569036 LAPPANG, FATIMAADJIVIR P
44 16 568367 LAZIM, NORLAILA N
45 16 535424 LOFRANCO, JERLIE MAE C
46 16 568381 LUCMAN, HANNAH M
47 16 568973 LUMA, MOHAMMAD NABEL D
48 16 568781 MACALANGCOM, AMINAH N
49 16 568704 MACARIMBANG, RAQUIB O
50 16 535244 MACASAYON, ABDUL QARNAIN P
51 16 569044 MACMAC, JAHARA D
52 16 568384 MACOD, ABDUL MALIK G
53 16 568323 MAMONGCARA, ASMAH M
54 16 568605 MANGCO, JALILAH B
55 16 568848 MANGODA, SITTIE RAIFAH M
56 16 568924 MANGONDAYA, AMERODIN
57 16 568313 MAUYAG, AIZZA FERRINA T
58 16 535582 MODESTO, SHERYL M
59 16 568642 MOHAMMAD, AKISA A
60 16 568629 MUTI, SUHEIMI S
61 16 535206 ORDEN, AILENE S
62 16 568577 OSOP, NAIDAH Z
63 16 568902 PALAO, ABDUL RAHMAN C
64 16 568451 PANGADAPUN, JANISAH G
65 16 568371 PANGANDAMAN, JANINA T
66 16 568488 POLAO, APIPAH D
67 16 568640 RANGIRIS, SALEM JR A
68 16 568811 RAUPAN, MHELHARRIE M
69 16 568679 SAMBITORY, SHAMINODEN M
70 16 568750 SAMPAO, NADJA L
71 16 568282 TAMPOGAO, NORJANAH M
72 16 568791 TANGGOTE, JUHAIRAH D
73 16 568408 USMAN, HANIA-PERSIA F
74 16 535136 USMAN, LADY ASREA A
75 16 568450 USMAN, SAMMER D
76 16 535710 USOP, ANWAR B
77 16 535273 UTRERA, JAYMAR V
***Nothing Follows***

Congratulations!

## October 2015 Civil Service Subprof Exam Result – CARAGA List of Passers

Below is the list of passers from CARAGA of the Civil Service Exam Subprofessional Examination conducted on October 18, 2015.

Seq. No. ExamNo Name
1 920792 ALFEREZ, ZSARAH MAGNE A
2 920355 ALIPANTE, APRILLE M
3 920418 AMOR, JOYCEL S
4 920214 AMORIO, LLENA MARIE G
5 920293 APAS, SIONNY R
6 920018 ARANCHADO, GRACE SHAMERGIE E
7 920246 AUTOR, ANDREW C
8 920726 BALATERO, ANECITO JR S
9 920402 BELTIS, KRISTER EVENZ M
10 920385 BISNAR, STACEY M
11 920243 BRAZA, NESTLIE MAY G
12 920690 BULLO, JESSAH MAE E
13 920227 BURDEOS, MICHAEL S
14 920165 CAPAROSO, MON JUSTIN CHRISTIAN V
15 920161 CARABUENA, ERWIN G
16 920630 CLERINO, LORWIN JADE T
17 920228 COTARES, CARMEL A
18 920038 DAGUPLO, SHERLY A
19 920183 DALAUTA, REYNARD V
20 920080 GALING, MONA PIA G
21 920817 GAMBLE, JHONALYN B
22 920300 GONZALES, JEFFERESON B
23 920317 ILLANA, NORMAN T
24 920119 JAPSON, CANDY CLAIRE C
25 920805 LAID, MARILOU V
26 920512 LARONG, HAZEL G
27 920481 LAVADOR, MARIANNE ROSE L
28 920193 LIPA, ELIZABETH C
29 920259 LOPEZ, MARA MAE B
30 920295 LUGTU, YVETTE CLAIRE L
31 920744 LUMANAO, EMMANUEL JR B
32 920555 LUSTIVA, JACKIE ALEJANDRO T
33 920592 MARIBAO, MARIZA B
34 920483 MIQUE, MARY CARRIE ANN A
35 920069 MONTIL, MARIA JOANNA N
36 920113 MONTON, JOSS ELMAR B
37 920359 MORDENO, JANE ROSE S
38 920365 NAVALES, BOYET P
39 920082 OCIO, JOCELYN L
40 920458 OTAZA, VELCHMEDA B
41 920026 PAREDES, MAY F
42 920576 PEREZ, SHARMAINE B
43 920044 SAID, SIGAY S
44 920620 SAMONTE, BERNA ROSE S
45 920652 SEMORLAN, LEONYL VIC P
46 920141 SIO, CRACIEL SHYN A
47 920463 TABUDLONG, ZAIDY L
48 920184 TAGARAO, MARIELLE M
49 920059 TIMOGAN, MARIBEL P
50 920276 TINAMPAY, SUZETTE T
51 920134 TUPAZ, EDRIEL ANJOE A
52 920188 UYGIOCO, IMMANUEL CHARLES M
53 920390 VILLOCIDO, JR D
54 920292 WAGA, ANA MYRNA ISABEL B
*** NOTHING FOLLOWS ***

Congratulations!

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