1. The document discusses properties and examples of quadratic equations. It covers topics like the sum and product of roots, symmetric and reciprocal roots, and the discriminant.
2. It provides examples of solving quadratic equations by finding values that satisfy certain properties of the roots. These include finding values for variables that would result in the roots summing to a given value or their product being equal to a given value.
3. The document also contains miscellaneous problems involving finding values of variables in quadratic equations given information about the roots, like if they are equal or reciprocal. It asks the reader to determine values that satisfy the given conditions.
Solucionario del examen de Admisión de la Universidad Nacional de Ingeniería de Matemáticas, tomado el 11/08/2014.
Desarrollado por la Academia Saco Oliveros
Solucionario del examen de Admisión de la Universidad Nacional de Ingeniería de Matemáticas, tomado el 11/08/2014.
Desarrollado por la Academia Saco Oliveros
Read| The latest issue of The Challenger is here! We are thrilled to announce that our school paper has qualified for the NATIONAL SCHOOLS PRESS CONFERENCE (NSPC) 2024. Thank you for your unwavering support and trust. Dive into the stories that made us stand out!
Students, digital devices and success - Andreas Schleicher - 27 May 2024..pptxEduSkills OECD
Andreas Schleicher presents at the OECD webinar ‘Digital devices in schools: detrimental distraction or secret to success?’ on 27 May 2024. The presentation was based on findings from PISA 2022 results and the webinar helped launch the PISA in Focus ‘Managing screen time: How to protect and equip students against distraction’ https://www.oecd-ilibrary.org/education/managing-screen-time_7c225af4-en and the OECD Education Policy Perspective ‘Students, digital devices and success’ can be found here - https://oe.cd/il/5yV
Model Attribute Check Company Auto PropertyCeline George
In Odoo, the multi-company feature allows you to manage multiple companies within a single Odoo database instance. Each company can have its own configurations while still sharing common resources such as products, customers, and suppliers.
How to Split Bills in the Odoo 17 POS ModuleCeline George
Bills have a main role in point of sale procedure. It will help to track sales, handling payments and giving receipts to customers. Bill splitting also has an important role in POS. For example, If some friends come together for dinner and if they want to divide the bill then it is possible by POS bill splitting. This slide will show how to split bills in odoo 17 POS.
2024.06.01 Introducing a competency framework for languag learning materials ...Sandy Millin
http://sandymillin.wordpress.com/iateflwebinar2024
Published classroom materials form the basis of syllabuses, drive teacher professional development, and have a potentially huge influence on learners, teachers and education systems. All teachers also create their own materials, whether a few sentences on a blackboard, a highly-structured fully-realised online course, or anything in between. Despite this, the knowledge and skills needed to create effective language learning materials are rarely part of teacher training, and are mostly learnt by trial and error.
Knowledge and skills frameworks, generally called competency frameworks, for ELT teachers, trainers and managers have existed for a few years now. However, until I created one for my MA dissertation, there wasn’t one drawing together what we need to know and do to be able to effectively produce language learning materials.
This webinar will introduce you to my framework, highlighting the key competencies I identified from my research. It will also show how anybody involved in language teaching (any language, not just English!), teacher training, managing schools or developing language learning materials can benefit from using the framework.
How to Create Map Views in the Odoo 17 ERPCeline George
The map views are useful for providing a geographical representation of data. They allow users to visualize and analyze the data in a more intuitive manner.
How to Make a Field invisible in Odoo 17Celine George
It is possible to hide or invisible some fields in odoo. Commonly using “invisible” attribute in the field definition to invisible the fields. This slide will show how to make a field invisible in odoo 17.
The French Revolution, which began in 1789, was a period of radical social and political upheaval in France. It marked the decline of absolute monarchies, the rise of secular and democratic republics, and the eventual rise of Napoleon Bonaparte. This revolutionary period is crucial in understanding the transition from feudalism to modernity in Europe.
For more information, visit-www.vavaclasses.com
Welcome to TechSoup New Member Orientation and Q&A (May 2024).pdfTechSoup
In this webinar you will learn how your organization can access TechSoup's wide variety of product discount and donation programs. From hardware to software, we'll give you a tour of the tools available to help your nonprofit with productivity, collaboration, financial management, donor tracking, security, and more.
The Indian economy is classified into different sectors to simplify the analysis and understanding of economic activities. For Class 10, it's essential to grasp the sectors of the Indian economy, understand their characteristics, and recognize their importance. This guide will provide detailed notes on the Sectors of the Indian Economy Class 10, using specific long-tail keywords to enhance comprehension.
For more information, visit-www.vavaclasses.com
The Art Pastor's Guide to Sabbath | Steve ThomasonSteve Thomason
What is the purpose of the Sabbath Law in the Torah. It is interesting to compare how the context of the law shifts from Exodus to Deuteronomy. Who gets to rest, and why?
4. Solución
1.
A) 4 B) −8 C) 2 D) −4 E) 8
Hallar “m”, si una raíz es – 2.
En la ecuación: x2 + 6x – m = 0
x2 + 6x – m = 0
(– 2)2 + 6 (– 2) – m = 0
4 – 12 – m = 0
– 8 – m = 0
– 8 = m Rpta. B
Reemplazando
5. Solución
2.
Rpta. B
A) – 1 B) 2 C) 1 D) 4 E) 8
x1 + x2 = − k + 7 = − 9
Calcular k para que la suma de raíces de:
sea igual a − 9.
x2
+ k + 8 x + 8 − x = 0
x2
+ k + 8 x + 8 − x = 0
x2 + k + 8 − 1 x + 8 = 0
x2 + k + 7 x + 8 = 0
−k − 7 = − 9
2 = k
𝑥1 + 𝑥2 =
−𝑏
𝑎
6. Solución
3.
Rpta. D
A) 6 B) 5 C) 4 D) 3 E) 2
x1 . x2
3 = n
Determinar n tal que el producto de raíces de la ecuación:
sea igual a 6. (n − 2)x2
−5x + 2n = 0
6
2
2
n
n
12
6
2
n
n
(n − 2)x2−5x + 2n = 0
12= 4n
𝑥1. 𝑥2 =
𝑐
𝑎
7. Solución
4.
Rpta. C
A) 14 B) 15 C) 16 D) 17 E) 18
x1 . x2
m = 16
Determinar m tal que el producto de raíces de la ecuación:
sea igual a 9.
2x2 + m − 1 x + m + 2 = 0
2x2 + m − 1 x + m + 1 = −1
9
2
2
m
2x2 + m − 1 x + m + 1 = −1
18
2
m
𝑥1. 𝑥2 =
𝑐
𝑎
8. A) 50 B) 100 C) 150 D) 200 E) 250
5.
Rpta. D
Solución
0
1
)
600
3
(
3 2
x
m
x
Hallar “m”, si la ecuación:
posee raíces simétricas.
0
3
600
3
m
0
600
3
m
600
3
m
200
m
𝑥1 + 𝑥2 = 0
𝑥1 + 𝑥2 =
Raíces simétricas:
9. A) 6 B) 7 C) 8 D) 9 E) 10
6.
Rpta. B
Solución
0
1
b
0
b
0
b
Hallar la mayor solución de la ecuación: x2 + bx + 2b - 49 = 0
si tiene raíces simétricas.
x2 − 49 = 0
(x+7)(x − 7) = 0
7
x
7
x
x2 + bx + 2b - 49 = 0
𝑥1 + 𝑥2 = 0
Raíces simétricas:
10. A) 6 B) 7 C) 8 D) 9 E) 10
7.
Rpta. E
Solución
Hallar “k”, si la ecuación: 0
)
9
(
7
)
1
2
( 2
k
x
x
k
posee raíces recíprocas.
1
1
2
9
k
k
1
2
9
k
k
k
10
𝑥1. 𝑥2 = 1
Raíces recíprocas:
𝑥1. 𝑥2 =
0
)
9
(
7
)
1
2
( 2
k
x
x
k
11. A) 1 B) 2 C) 3 D) 4 E) 5
8.
Rpta. A
Solución
1
2
4
a
a
2
4
a
a
a
1
Dada la ecuación: (a+2)x2 – 10x + 4 – a = 0. Hallar el valor
de “a” si tiene raíces reciprocas.
a
2
2
𝑥1. 𝑥2 = 1
Raíces recíprocas:
𝑥1. 𝑥2 =
(a+2)x2 – 10x + 4 – a = 0
12. A) 2 B) 3 C) 4 D) 5 E) 6
9.
Rpta. C
Solución
1
4
k
4
k
4
x
Determine una raíz entera de la ecuación cuadrática:
kx2 − 17x + 4 = 0 que posee raíces reciprocas.
(x− 4)(4x − 1) = 0
4x2 − 17x + 4 = 0
4
/
1
x
𝑥1. 𝑥2 = 1
Raíces recíprocas:
kx2 − 17x + 4 = 0
13. Solución
10.
Rpta. A
A) 12 B) 11 C) 10 D) 9 E) 8
Determinar el valor positivo de k de modo que las dos
raíces de la ecuación x2 − kx + 36 = 0 sean iguales.
b2 − 4ac = 0
(− k)2 − 4 ·1·36 = 0
k2 =144
∆= b2 − 4ac = 0
Raíces iguales:
x2 − kx + 36 = 0
14. Solución
11.
Rpta. E
A) 9 B) 8 C) 7 D) 6 E) 5
(− 2k)2 − 4 ·1·(8k− 15) = 0
4(k2 − 8k +15) = 0
Hallar “k” si la ecuación: x2 − 15 − k 2x − 8 = 0
Tiene raíces iguales.
x2 − 15 − k 2x − 8 = 0 x2 − 15 − 2kx + 8k = 0
x2 − 2kx + 8k − 15 = 0
4k2 − 32k +60 = 0
(k − 3)(k − 5) = 0
k =3 k =5
∆= b2 − 4ac = 0
Raíces iguales:
15. Solución
12.
Rpta. C
A) 8 B) 7 C) 6 D) 5 E) 4
(− 2m)2 − 4 ·(m+3)·4 = 0
m = − 2
m = 6
Halle el mayor valor de “m” para que la ecuación:
m + 3 x2
− 2mx + 4 = 0, tenga una única solución.
4m2 − 16m − 48 = 0
4(m2 − 4m − 12) = 0
(m − 6)(m + 2) = 0
m + 3 x2 − 2mx + 4 = 0
∆= b2 − 4ac = 0
Raíces iguales:
16. Solución
13.
Rpta. D
A) 44 B) 33 C) 22 D) 11 E) 88
(− 8)2 − 4 ·1·n = 20
Para que valor de n el discriminante de la ecuación:
es igual a 20. x2 − 8x + n = 0
x2 − 8x + n = 0
∆= b2 − 4ac = 20
64− 4n = 20
44 = 4n
11 = n
17. Solución
14.
Rpta. B
A) 3 B) 4 C) 5 D) 6 E) 7
0
1
5
2 2
x
x
Siendo Las raíces de:
b
a
E
1
1
Hallar:
"
"
"
" b
a
ab
b
a
E
b
a
E
1
1
2
5
b
a
2
1
.
b
a
5
ab
b
a
18. Solución
15.
Rpta. A
A) 16 B) 15 C) 14 D) 13 E) 12
Siendo Las raíces de:
a
b
b
a
E
Hallar:
"
"
"
" b
a
b
a
E
1
1
3
2
6
b
a
2
1
.
b
a
0
1
6
2 2
x
x
ab
b
a
E
2
2
2 2 2
a b a b 2ab
16
2
2
ab
b
a
E
( 3)2 = a2 +b2 +2(1/2)
a2 +b2 = 9 −1
a2 +b2 = 8
20. Solución
1.
Rpta. D
A) 4/3 B) 7/4 C) 1/2 D) 4/7 E) 3/7
Si en la ecuación: x2 – 5ax + 3a = 0; una de las raíces
es 2. Indicar el valor que adopta “a”.
x2 – 5ax + 3a = 0
(2)2 – 5a(2) + 3a = 0
4 – 10a + 3a = 0
4 – 7a = 0
4/7 = a
21. Solución
2.
A) 26/9 B) 52/9 C) 6 D) 52/3 E) 26/3
Si a y b son las raíces de la ecuación : 3x2 + 2x − 4 = 0
Calcule el valor de: (a − b)2
3
2
b
a
3
4
.
b
a
2 2
a b a b 4ab
(− 2/3)2 − (a − b)2 = 4(− 4/3)
52/9= (a − b)2
4/9 − (a − b)2 = − 16/3)
4/9 + 16/3 = (a − b)2
Rpta. B
LEGENDRE
Raíces a y b
22. Solución
3.
Rpta. D
A) 10 B) 20 C) −20 D) −15 E) 15
Si a y b son las raíces de la ecuación: x2 + mx + 36 = 0
Tal que (1/a) + (1/b) = 5/12 Calcular el valor de: “m”
b
a
1
1
12
5
ab
b
a
12
5
36
m
12
5
180
12
m
15
m
𝑥1 + 𝑥2 =
−𝑏
𝑎
𝑥1. 𝑥2 =
𝑐
𝑎
36
m
23. Solución
4.
Rpta. D
A) 3 B) 4 C) 5 D) 6 E) 7
3
2
6
b
a
2
3
.
b
a
( a+ b)2 = a2 +b2 +2ab
( − 3)2= a2 +b2 +2(3/2)
a2 +b2 = 9 − 3
a2 +b2 = 6
Sea el conjunto solución de la ecuación: (x2 /3) + x = − 1/2
es CS = {a ; b}. Indique el valor numérico de “a2 + b2”
2x2 + 6x + 3 = 0
Multiplicando MCM=6 BINOMIO
Raíces:
24. Solución
5.
Rpta. D
A) 14/3 B) 17/4 C) 17/2 D) 35/8 E) 35/4
Si la ecuación: 𝐾𝑥2
+ 2𝐾 + 1 𝑥 + 𝐾 = 0 , tiene raíces iguales,
hallar el producto de las raíces de la siguiente ecuación:
4𝐾 + 3 𝑦2
+ 3𝐾𝑦 − 4𝐾2
+ 9 = 0
𝐾𝑥2
+ 2𝐾 + 1 𝑥 + 𝐾 = 0
(2k+ 1)2 − 4 ·K·K = 0
4k2 +4k +1 − 4𝑘2
= 0 k=− 1/4
4𝐾 + 3 𝑦2 + 3𝐾𝑦 − 4𝐾2 + 9 = 0
x1 . x2 =
−4𝐾2+9
4𝑘+3 x1 . x2 =
−4
1
16
+ 9
4 −
1
4
+ 3
=
35
4
2
= 35/8
25. Solución
6.
Rpta. D
A) 1 B) 2 C) 3 D) 4 E) 5
Para que valor de “k”,
las raíces de la ecuación: 1
k
1
k
2
x
5
x
3
x2
Son simétricas
1
k
1
k
2
x
5
x
3
x2
(k+1)(x2 + 3x)= (k − 1)(5x+2)
(k+1)x2 + (3k+3)x= (5k− 5)x+(2k− 2)
(k+1)x2 + (− 2k+8)x + 2− 2k = 0
0
1
8
2
k
k 0
8
2
K
K
4
26. A) 7 B) 6 C) 5 D) 4 E) 3
7.
Rpta. E
Solución
1
2
4
a
a 2
4
a
a
a
1
Dada la ecuación: (a+2)x2 −10x + 4 − a = 0. Hallar la
mayor raíz si tiene raíces reciprocas.
a
2
2
3x2 − 10x + 3 = 0
(x− 3)(3x − 1) = 0
3
x
3
/
1
x
27. Solución
8.
A) 11 B) 12 C) 13 D) 16 E) 26
b
b
b
a
1
)
(
30
1
30
.
b
a
2 2
a b a b 4ab
(b)2 − ( 7 )2 = 4(30)
b = 13 Rpta. C
LEGENDRE
Sabiendo que las raíces de la ecuación en “x” : 𝑥2
−𝑏𝑥 + 30 = 0,
son positivas y la diferencia entre ellas es 7, halle el valor de “b”.
7
b
a
b2 − 49 = 120
b2 = 169
28. Solución
9.
Rpta. A
A) – 1 B) 2 C) 1 D) –2 E) 3
En la ecuación: x2 – (m + n)x + 2m + 2 = 0 tiene por
raíces a x1 = 2 y x2 = 3 Hallar: “m – n”
m + n = x1 + x2 = 2 + 3 = 5
x1 . x2 = 2m + 2
x1 + x2 = m + n
2m + 2 = x1 . x2 = 2.3 = 6
m + n = 5
2m + 2 = 6
m = 2
n = 3
m – n = – 1
29. Solución
10. En la siguiente ecuación:
hallar la Suma de raíces.
Rpta. D
A) 2 B) – 2 C) 4 D) – 4 E) 3
x(x + 2) + 5 = 3(2 – x) + x – 4
x(x + 2) + 5 = 3(2 – x) + x – 4
x2 + 2x + 5 = 6 – 3x + x – 4
x2 + 4x + 3 = 0
Suma de raíces = – 4
30. Solución
11.
Rpta. B
A)−1/3 B) 1/3 C) 1/2 D) − 1/2 E) 3/2
Siendo Las raíces de:
n
m
E
1
1
Hallar:
"
"
"
" n
m
mn
n
m
E
n
m
E
1
1
2
1
n
m
2
3
.
n
m
3
/
1
mn
n
m
0
3
2 2
x
x
𝑥1 + 𝑥2 =
−𝑏
𝑎
𝑥1. 𝑥2 =
𝑐
𝑎
31. Solución
12.
Rpta. B
A) 7 B) 8 C) 9 D) 10 E) 11
b2 − 4ac = 0
(k− 2)2 − 4 ·9·1 = 0
(k− 2)2 =36
Hallar el mayor valor de “k” para que la ecuación:
9x2 + (k-2)x + 1 = 0 admita solución única.
k = − 4
k = 8
32. Solución
13.
Rpta. C
A) 4 B) 11/4 C) − 9/4 D) − 11/4 E) 9
Sea la ecuación: 2x2 +3x +5 =0, de raíces a ; b Hallar “a2 +b2”
2
3
b
a
2
5
.
b
a
( a+ b)2 = a2 +b2 +2ab
( − 3/2)2=a2 +b2 +2(5/2)
a2 +b2 =(9/4)− 5
a2 +b2 = − 11/4