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Without an equality sign the expression given can't be considered to be an equation.

Q: What equations is the result of completing the square on x2 - 6x - 9 0?

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If you mean: 6x-3y = -33 and 2x+y = -1 Then solving the simultaneous equations by substitution: x = -3 and y = 5

If the square root of 6x-2=4 than we can square both sides and we have 6x-2=16 or 6x=18 and the solution set is the set that contains the number 3.

I assume you mean the equation 2x2 + 12x - 10 = 0. Divide by 2: x2 + 6x - 5 = 0 To complete the square, the constant part (the part with no variable) must be half the linear part (1/2 of 6), squared, so in this case you add 14 to each side: x2 + 6x +9 = 14 This can be factored: (x+3)2 = 14 Now, just take the square root of each side: x + 3 = (plus-or-minus) 14 Solve the resulting two equations (one for plus, one for minus).

√(36x4) is written as: √((6x²)²) Then, this is evaluated to 6x².

Thanks to limitations of the browser used for posting questions, it is not possible to tell whether the sides are 6x - 7 units or 6x + 7 units. For 6x - 7 units, area = 36x2 - 84x + 49 square units. For 6x + 7 units, area = 36x2 + 84x + 49 square units.

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y = x2 - 6x + 2 y = x2 - 6x + 9 - 7 y = (x - 3)2 - 7

x2 + 6x = 7 ⇒ x2 + 6x + 9 = 7 + 9 ⇒ (x + 3)2 = 16 ⇒ x + 3 = ±4 ⇒ x = -7 or 1

x^2 - 6x - 2 = 0 x^2 - 6x = 2 x^2 - 6x + 9 = 2 + 9 = 11 (x - 3)^2 = 11 x - 3 = +/- sqrt(11) x = 3 +/- sqrt(11)

This quadratic equation has no solutions because the discriminant is less than zero.

If: 3x^2 = 6x+4 Then: 3^2 -6x-4 = 0 Divide all terms by 3: x^2 -2x-4/3 = 0 Completing the square: (x-1)^2 -7/3 = 0 => (x-1)^2 = 7/3 or as 21/9 If: (x-1)^2 = 21/9 Then: x = 1 +/- (square root 21) over 3 which are the exact solutions

Expression. Equations have equals signs.

If you mean: 6x-3y = -33 and 2x+y = -1 Then solving the simultaneous equations by substitution: x = -3 and y = 5

If the square root of 6x-2=4 than we can square both sides and we have 6x-2=16 or 6x=18 and the solution set is the set that contains the number 3.

You mean find vertex. Solving for X is another matter, using the quadratic formula here.Vertex form:X2 + 6X - 2 = 0X2 + 6X = 2Halve the linear term( 6), square it and add it to both sidesX2 + 6X + 9 = 2 + 9factor the left side and gather terms on the right(X + 3)2 = 11(X + 3)2 - 11 = 0================vertex form(- 3, - 11 )==========Vertex

√(36x4) is written as: √((6x²)²) Then, this is evaluated to 6x².

I assume you mean the equation 2x2 + 12x - 10 = 0. Divide by 2: x2 + 6x - 5 = 0 To complete the square, the constant part (the part with no variable) must be half the linear part (1/2 of 6), squared, so in this case you add 14 to each side: x2 + 6x +9 = 14 This can be factored: (x+3)2 = 14 Now, just take the square root of each side: x + 3 = (plus-or-minus) 14 Solve the resulting two equations (one for plus, one for minus).

Thanks to limitations of the browser used for posting questions, it is not possible to tell whether the sides are 6x - 7 units or 6x + 7 units. For 6x - 7 units, area = 36x2 - 84x + 49 square units. For 6x + 7 units, area = 36x2 + 84x + 49 square units.