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3 years ago

What is the solution to the equation?

Mathematics

2 answers:

pashok25 [27]3 years ago

4 0

Answer:

x= 27

Step-by-step explanation:

Move <u>constant</u> to the right-hand side and change its sign.

$\sqrt{2x+10}$ = 2+6

Add the numbers.

$\sqrt{2x+10}$ = 8

Square both sides of the equation.

2x+10= 64

Move the <u>constant</u> to the right-hand side and change its sign.

2x= 64-10

Subtract the numbers.

2x= 54

Divide both sides of the equation by 2.

x= 27

Mamont248 [21]3 years ago

3 0

Add 6 to 2

_/2x+10 = 8

Square both sides

2x + 10 = 64

Subtract 10

2x = 54

Divide two

X = 27

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Step-by-step explanation:

$\sqrt{2x+4}-\sqrt{x}=2$

Isolating √(2x+4): Addind √x both sides of the equation:

$\sqrt{2x+4}-\sqrt{x}+\sqrt{x}=2+\sqrt{x}\\ \sqrt{2x+4}=2+\sqrt{x}$

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Simplifying on the left side, and applying on the right side the formula:

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$2x+4=(2)^{2}+2(2)(\sqrt{x})+(\sqrt{x})^{2}\\ 2x+4=4+4\sqrt{x}+x$

Isolating the term with √x on the right side of the equation: Subtracting 4 and x from both sides of the equation:

$2x+4-4-x=4+4\sqrt{x}+x-4-x\\ x=4\sqrt{x}$

Squaring both sides of the equation:

$(x)^{2}=(4\sqrt{x})^{2}\\ x^{2}=(4)^{2}(\sqrt{x})^{2}\\ x^{2}=16 x$

This is a quadratic equation. Equaling to zero: Subtract 16x from both sides of the equation:

$x^{2}-16x=16x-16x\\ x^{2}-16x=0$

Factoring: Common factor x:

x (x-16)=0

Two solutions:

1) x=0

2) x-16=0

Solving for x: Adding 16 both sides of the equation:

x-16+16=0+16

x=16

Let's prove the solutions in the orignal equation:

1) x=0:

$\sqrt{2x+4}-\sqrt{x}=2\\ \sqrt{2(0)+4}-\sqrt{0}=2\\ \sqrt{0+4}-0=2\\ \sqrt{4}=2\\ 2=2$

x=0 is a solution

2) x=16

$\sqrt{2x+4}-\sqrt{x}=2\\ \sqrt{2(16)+4}-\sqrt{16}=2\\ \sqrt{32+4}-4=2\\ \sqrt{36}-4=2\\ 6-4=2\\ 2=2$

x=16 is a solution

Then the solutions are x=0 and x=16

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