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3 years ago

Which of these equations have no solution? check all that apply

Mathematics

2 answers:

finlep [7]3 years ago

8 0

Answer:

A, B, and E

Step-by-step explanation:

Let's look through each of the answer choices.

A) 2(x + 2) + 2 = 2(x + 3) + 1

Simplify this by expanding out the parentheses:

2x + 2 * 2 + 2 = 2x + 2 * 3 + 1

2x + 4 + 2 = 2x + 6 + 1

2x + 6 = 2x + 7

6 = 7

Well, we know that 6 is NOT equal to 7, so this has no solutions.

B) 2x + 3(x + 5) = 5(x – 3)

Simplify this by expanding out the parentheses:

2x + 3x + 3 * 5 = 5x - 5 * 3

5x + 15 = 5x - 15

15 = -15

Again, we know that 15 is NOT equal to -15, so this also has no solutions.

C) 4(x + 3) = x + 12

Simplify this by expanding out the parentheses:

4x + 4 * 3 = x + 12

4x + 12 = x + 12

3x = 0

x = 0

This actually does have a solution when x = 0, so eliminate this.

D) 4 – (2x + 5) = (–4x – 2)

Simplify this by expanding out the parentheses:

4 - 2x - 5 = -4 - 2

-2x - 1 = -6

-2x = -5

x = 2.5

This also has a solution when x = 2.5, so eliminate this.

E) 5(x + 4) – x = 4(x + 5) – 1

Simplify this by expanding out the parentheses:

5x + 5 * 4 - x = 4x + 4 * 5 - 1

4x + 20 = 4x + 19

20 = 19

Here, we know that 20 is NOT equal to 19, so this has no solution.

Hope this helps!

Lemur [1.5K]3 years ago

5 0

Answer:

First, second and fifth

Step-by-step explanation:

2(x + 2) + 2 = 2(x + 3) + 1

2x + 6 = 2x + 7

6 = 7 (false)

2x + 3(x + 5) = 5(x – 3)

5x + 15 = 5x - 15

15 = -15 (false)

4(x + 3) = x + 12

4x + 12 = x + 12

3x = 0

x = 0

4 – (2x + 5) = (–4x – 2)

4 - 2x - 10 = -4x - 2

2x = 4

x = 2

5(x + 4) – x = 4(x + 5) – 1

4x + 20 = 4x + 19

20 = 19 (false)

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Step-by-step explanation:

We need to solve by factoring: $x^4-8x^2+12=0$

Let x^2 = u then x^4 = u^2

Putting values:

$u^2-8u+12=0$

Factorizing:

$u^2-6u-2u+12=0$

$u(u-6)-2(u-6)=0$

$(u-2)(u-6)=0$

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Putting value of u=x^2

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Taking square root:

$x=\pm \sqrt{2}\,\,and\,\,x=\pm \sqrt{6}$

$x=\sqrt{2}\,\,,x=-\sqrt{2}\,\,,x=\sqrt{6}\,\,,x=-\sqrt{6}$

So, After solving by factoring $x^4-8x^2+12=0$ we get $x=\sqrt{2}\,\,,x=-\sqrt{2}\,\,,x=\sqrt{6}\,\,,x=-\sqrt{6}$

Keywords: Solving Quadratic Equation

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scZoUnD [109]

Answer: $H_0: \sigma=1.8$

Step-by-step explanation:

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