How to complete the square? Grade 10 math question?

Question

y=3x^2-24x+10

Answer 1

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divide each number by 3.

x^2-8x+10/3

x^2-8x=-10/3

b/2=-8/2=-4

(b/2)^2=16

(x-4)^2=-10/3+16

(x-4)^2=38/3

x-4=(+/-)sqrt(38/3)

x=(+/-)sqrt(38/3)+4.

Answer 2

Since we know the following is true:

(x + y)² = x² + 2xy + y²

Then if we have something that is :

x² + bx

Then we can calculate a constant to add to it that makes it fit the pattern above, so you can factor them to a binomial squared.

To do that, you take half of the coefficient of x (my b shown above), then square it.

So the general form of:

x² + bx

would become:

x² + bx + (b² / 4)

This then factors to :

(x + b/2)²

Now for using this information on your example:

y = 3x² - 24x + 10

Since you are usually solving these for when y = 0, I'll change the equation to:

3x² - 24x + 10 = 0

Now we need to move the constant over to the right side, then make it so the coefficient of x² is 1

3x² - 24x = -10 (subtract 10 from both sides)

x² - 8x = -10/3 (divide 3 from both sides)

Now we have the form we need, to "complete the square", we have to find out what we have to add to both sides so we can factor the left half. As I stated before, half the coefficient of x, then square it.

Since we have -8 for a coefficient, halving it gives -4, then squaring it leaves 16. So add 16 to both sides:

x² - 8x + 16 = -10/3 + 16

x² - 8x + 16 = -10/3 + 48/3

x² - 8x + 16 = 38/3

Now factor the left half of the equation:

(x - 4)² = 38/3

Now take the square root of both halves:

x - 4 = ±√(38/3)

Add 4 to both sides:

x = 4 ± √(38/3)

Rationalize the denominator:

x = 4 ± √(38 * 3) / 3

x = 4 ± √(114) / 3

Answer 3

3(x^2-8)+10 =0

3x^2-24 =-10 divide by 3

x^2-8= -10/3 add to each side (-8/2)^2=16

x^2 -8 + 16= -10/3 + 16

(x -4)^2=38/3

If you want to solve it take square root of both sides

(x-4)=±sqrt(38/3)

x=4±sqrt(38/3)

Answer 4

y = 3(x² - 8x) + 10

y = 3(x² - 8x + 16) - 48 + 10

y = 3(x - 4)² - 38

Answer 5

divide by potential of 6 h^2-h/6=one million/3 h^2-h/6+one million/one hundred forty four=one million/3+one million/one hundred forty four (h-one million/12)^2=40 8/one hundred forty four+one million/one hundred forty four (h-one million/12)^2=40 9/one hundred forty four h-one million/12=7/12 h=8/12=2/3 h-one million/12=-7/12 h=-6/12=-one million/2 h=2/3 and h=-one million/2 greater straightforward to factor it 6h^2 - h - 2 =0 aspects of -12 which upload as much as -one million -4 and 3 6h^2+3h-4h-2 3h*(2h+one million)-2*(2h+one million)=0 (2 h+one million) (3 h-2) = 0 h=-one million/2 h=2/3