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Completando el TCP

Completando el TCP

Assessment

Presentation

Mathematics

12th Grade

Practice Problem

Medium

Created by

Michelle B.M.

Used 1+ times

FREE Resource

1 Slide • 9 Questions

1

2

Multiple Choice

Paso 1 Colocar el término independiente de lado derecho de la igualdad 2x213x+15=02x^2-13x+15=0

1

2x213x+15=02x^2-13x+15=0

2

2x213x=+152x^2-13x=+15

3

2x213x=152x^2-13x=-15

3

Multiple Choice

Paso 2 Factoriza el coeficiente del término principal

2x213x=152x^2-13x=-15

1

2x213x+15=02x^2-13x+15=0

2

2(x213)x=+152\left(x^2-13\right)x=+15

3

2(x2132x)=152\left(x^2-\frac{13}{2}x\right)=-15

4

Multiple Choice

Paso 3 Obtener el tercer término del trinomio cuadrado perfecto

2(x2132x)=152\left(x^2-\frac{13}{2}x\right)=-15

1

(132÷2)2\left(\frac{13}{2}\div2\right)^2

2

(132÷2)2\left(-\frac{13}{2}\div2\right)^2

3

(132)2\left(\frac{13}{2}\right)^2

5

Multiple Choice

Paso 4 Completa el trinomio cuadrado perfecto sin alterar la igualdad

2(x2132x)=152\left(x^2-\frac{13}{2}x\right)=-15

1

2(x2132x+16916)=15+2(16916)2\left(x^2-\frac{13}{2}x+\frac{169}{16}\right)=-15+2\left(\frac{169}{16}\right)

2

2(x2132x+1692)=15+2(1692)2\left(x^2-\frac{13}{2}x+\frac{169}{2}\right)=-15+2\left(\frac{169}{2}\right)

3

2(x2+132x1694)=15+2(16916)2\left(x^2+\frac{13}{2}x-\frac{169}{4}\right)=-15+2\left(\frac{169}{16}\right)

6

Multiple Choice

Paso 5 Factoriza el trinomio cuadrado perfecto sin alterar la igualdad

2(x2132x+16916)=15+2(16916)2\left(x^2-\frac{13}{2}x+\frac{169}{16}\right)=-15+2\left(\frac{169}{16}\right)

1

2(x134)2=15+2(16916)2\left(x-\frac{13}{4}\right)^2=-15+2\left(\frac{169}{16}\right)

2

2(x2+134x)2=15+2(16916)2\left(x^2+\frac{13}{4}x\right)^2=-15+2\left(\frac{169}{16}\right)

3

2(x2+132)2=15+2(16916)2\left(x^2+\frac{13}{2}\right)^2=-15+2\left(\frac{169}{16}\right)

7

Multiple Choice

Paso 6 Simplificar la expresión de la derecha de la igualdad

2(x2134)2=15+2(16916)2\left(x^2-\frac{13}{4}\right)^2=-15+2\left(\frac{169}{16}\right)

1

15+2(16916)=4916-15+2\left(\frac{169}{16}\right)=\frac{49}{16}

2

15+2(16916)=3916-15+2\left(\frac{169}{16}\right)=\frac{39}{16}

3

15+2(16916)=498-15+2\left(\frac{169}{16}\right)=\frac{49}{8}

8

Multiple Choice

Paso 7 Despeja el binomio cuadrado

2(x2134)2=4982\left(x^2-\frac{13}{4}\right)^2=\frac{49}{8}

1

(x134)2=988\left(x-\frac{13}{4}\right)^2=\frac{98}{8}

2

(x134)2=4916\left(x-\frac{13}{4}\right)^2=\frac{49}{16}

3

(x134)2=498\left(x-\frac{13}{4}\right)^2=\frac{49}{8}

9

Multiple Choice

Paso 8 Aplica raíz cuadrada de ambos lados de la igualdad y despeja x

(x134)22=±49162\sqrt[2]{\left(x-\frac{13}{4}\right)^2}=\pm\sqrt[2]{\frac{49}{16}}

1

x=±74134x=\pm\frac{7}{4}-\frac{13}{4}

2

x=±134+4916x=\pm\frac{13}{4}+\frac{49}{16}

3

x=±74+134x=\pm\frac{7}{4}+\frac{13}{4}

10

Multiple Choice

Paso 9 Los valores de x son

x=±74+134x=\pm\frac{7}{4}+\frac{13}{4}

1

x1=5

x2=3/2

2

x1=-5

x2=2/3

3

x1=-2

x2=-5/4

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