# Can you solve 2x^2 = 200 in the method completing the square

## Answers

Hello there ^ _ ^

2x² = 200

We need to start by dividing both sides by 2

2x²/2 = 200/2

x² = 100

Now we can take square root.

x = +/-√100

x = 10 or x = -10

I hope this help!

Hello there ^ _ ^

2x² = 200

We need to start by dividing both sides by 2

2x²/2 = 200/2

x² = 100

Now we can take square root.

x = +/-√100

x = 10 or x = -10

I hope this help!

That work is right so thx for helping her or him

That work is right so thx for helping her or him

(a) y = 3(x^2 -4x) +17 = 3(x^2 -4x +4) +17 -3*4

y = 3(x -2)^2 +5

The turning point at (2, 5) is a minimum.

(b) y = -5(x^2 -8x) -70 = -5(x^2 -8x +16) -70 +80

y = -5(x -4)^2 +10

The turning point at (4, 10) is a maximum.

[tex]Use the method completing the square to write each of the following functions in the form y=a(x-h)^2[/tex]

x = -1 ± √6 / 5

Step-by-step explanation:

Solve [tex]5x^{2}[/tex] + 2x = 1

move 1 to the left side of the equation by subtracting it from both sides.

[tex]5x^{2}[/tex] + 2x - 1 = 0

Use the quadratic formula to find the solutions.

-b± [tex]\sqrt{b^2 - 4 (ac)}[/tex]/ 2a

Substitute the values a = 5 , b = 2, and c = -1 into the quadratic formula and solve for x

-2±[tex]\sqrt{2^2 - 4(5 - 1)}[/tex]/ 2 ⋅ 5

Simplify.

x = -2 ± 2 [tex]\sqrt{6}[/tex]/ 10

x = -1 ± [tex]\sqrt{6}[/tex]/ 5

Exact Form:

x = -1 ± [tex]\sqrt{6}[/tex]/ 5