Question

ES URGENTE!!! POR FAVOR AYUDA!!!

3. Resuelve las siguientes ecuaciones cuadráticas.

4x2=0
4x2-3x=0
x2+5x+6=0
n2-5n+4=0
3x2+x-2=0
6x2+3x-3=0

4. Dado el siguiente rectángulo, determina cuál debe ser el valor de x que permita obtener el área que se indica.

Answer 1

Answer:

• 1. x = 0

• 2. x = 0 and x = 3/4

• 3. x = - 2 and x = - 3

• 4. n = 4 and n = 1

• 5. x = -1 and x = 2/3

• 6. x = -1 and x = 1/2

• 7. Problem 4: x = 10 cm

Explanation:

1. 4x²=0

a) Divide both sides by 4:

• x² = 0

b) Take square root of both sides

• x = 0

c) Solution: there is only one solution: x = 0.

d) Substitute to verify:

• 4(0)² = 0

• 4(0) = 0

• 0 = 0 $\checkmark$

2. 4x² - 3x = 0

a) Common factor x:

• x(4x - 3) = 0

b) Zero product property: any or both factors must equal zero

i) x = 0

ii) 4x - 3 = 0

   4x = 3

     x = 3/4

Solutions x = 0 and x = 3

c) Verify the solutions:

i) x = 0 ⇒ 4(0)² - 3(0) = 0 - 0 = 0 $\checkmark$

ii) x = 3/4 ⇒ 4(3/4)² - 3(3/4) =3²/4 - 9/4 = 9/4 - 9/4 = 0 $\checkmark$

3. x² + 5x + 6 = 0

a) Factor: find two numbers whose sum is 5 and product is 6: 3 and 2:

• (x + 2)(x + 3) = 0

b) Zero product property:

i) x  + 2 = 0

  x = - 2

ii) x + 3 = 0

   x = - 3

   

The two solutions are x = - 2 and x = -3

c) Verify

i) x = - 2 ⇒ (-2)² + 5(-2) + 6 = 4 - 10 + 6 = - 6 + 6 = 0 $\checkmark$

ii) x = - 3 ⇒ (-3)² + 5(-3) + 6 = 9 - 15 + 6 = 9 - 9 = 0 $\checkmark$

4. n² - 5n + 4 = 0

a) Factor: find two numbers whose sum is -5 and product is 4: -4 and -1

• (n - 4)(n - 1) = 0

b) Zero product property:

i) n - 4 = 0 ⇒ n = 4

ii) n - 1 = 0 ⇒ n = 1

The two solutions are n = 4 and n = 1

c) Verify the solutions:

i) n = 4 ⇒(4)² - 5(4) + 4 = 16 - 20 + 4 = 16 - 16 = 0 $\checkmark$

ii) n = 1 ⇒ (1)² - 5(1) + 4 = 1 - 5 + 4 = 1 - 1 = 0$\checkmark$

5. 3x² + x - 2 = 0

a) Substitute x with 3x - 2x

• 3x² + 3x - 2x - 2 = 0

b) Group the terms (associative property)

• (3x² + 3x) - (2x + 2)

c) Factor each group:

• 3x(x + 1) - 2(x + 1) = 0

d) Common factor x + 1:

• (x + 1) (3x - 2) = 0

e) Zero product property

i) x + 1 = 0 ⇒ x = - 1

ii) 3x - 2 = 0 ⇒ x = 2/3

The two solutions are x = -1 and x = 2/3

f) Verify the solutions:

i) x = - 1 ⇒ 3(-1)² + (-1) - 2 = 3 - 1 - 2 = 2 - 2 = 0 $\checkmark$

ii) x = 2/3 ⇒ 3(2/3)² + (2/3) - 2 = 4/3 + 2/3 - 2 = 6/3 - 2 = 0 $\checkmark$

6. 6x² + 3x - 3 = 0

a) Divide both side by 3:

• 2x² + x - 1 = 0

b) Substiture x with 2x - x

• 2x² + 2x - x - 1 = 0

c) Group terms:

• (2x² + 2x) - (x + 1) = 0

d) Factor each binomial

• 2x(x + 1) -  (x + 1) = 0

e) Common factor x + 1

• (x + 1)(2x - 1) = 0

f) Zero product property

i) x + 1 = 0 ⇒ x = - 1

ii) 2x + 1 = 0

     x = 1/2

The two solutions are x = - 1 and x =  1/2

e) Verify

i) x = - 1 ⇒ 6(-1)² + 3(-1) - 3 = 6 - 3 - 3 = 0 $\checkmark$

ii) x = 1/2 ⇒ 6(1/2)² + 3(1/2) - 3 = 6/4 + 3/2 - 3 = 3 - 3 = 0 $\checkmark$

7. Problem

4. Dado el siguiente rectángulo, determina cuál debe ser el valor de x que permita obtener el área que se indica.

The area of a rectangle is the product of the length and the width:

• Area = (x - 4)(x + 1) = 66

Solve the equation:

a) Distributive property:

• x² + x - 4x - 4 = 66

b) Add like terms:

• x² -3x - 4 = 66

c) Subtract 66 from both sides

• x² -3x - 70 = 0

d) Factor: find two numbers whose sum is -3 and product is - 70: - 10 and 7

• (x - 10)(x + 7) = 0

e) Zero product property:

i) x - 10 = 0 ⇒ x = 10

ii) x + 7 = 0 ⇒ x = -7

x = - 7 is not a valid solution because that would make the side lengths negative.

Thus, the only solution is x = 10 cm

f) Verify:

• (10 - 4)(10 + 1) = 6 × 11 = 66 cm² $\checkmark$