Math

Solve each of the following quadratic equations by completing the aquare.

1. x^2-6x-3=0

2. 2x^2+10x+11=0

Step 1.
Move the constant to the other side.
X^2 -6x + ?? = 3 + ??
To find ??, take 1/2 the x term and square it, then add it to both sides; therefore, 1/2*6 = 3 and 3^2 = 9.
X^2 -6x + 9 = 3 + 9
X^2 -6x + 9 = 12.
(X+3)^2 = 12
X+3 = sq rt 12
Solve for X. Post your work if you get stuck. The other one is worked the same way.

  1. 👍
  2. 👎
  3. 👁

Respond to this Question

First Name

Your Response

Similar Questions

  1. Algebra

    1.Solve the system of equations. y = 2x^2 - 3 y = 3x - 1 a. no solution b. (-1/2, 5), (2, -5/2) c. (-1/2, -5/2), (2,5) d. (1/2, 5/2), (2, 5) 2.how many real number solutions does the equation have 0 = -3x^2 + x - 4 a. 0 b. 1 c. 2

  2. Math

    Consider the general quadratic equation ax^2 + bx + c = 0, where a is not equal to 0. Solve for x by completing the square.

  3. Algebra

    equations and inequalities unit test for questions 1-5, solve the equation 4. 3(4-2x)=-2x (1 point) a. -1*** b. 1 c. 2 d. 3 5. 5.4+0.2x=-1.4x+8.6 a. 2 b. -2 c. .2*** d. 20 For questions 6-7, simplify the expression 6. -4 - 5(x+8)

  4. Algebra Question

    Quadratic equations can be solved by graphing, using the quadrat completing the square, and factoring. What are the pros and cons these methods? When might each method be most appropriate?

  1. linear systems

    The statement that is false is A. A system of quadratic-quadratic equations can have exactly one solution. B. A system of quadratic-quadratic equations has no solutions if the graphs do not intersect. C. It is impossible for a

  2. Math

    Lesson 8: Systems of Linear and Quadratic Equations Check my work 1. Solve the system of equations. y = 2x^2 - 3 y = 3x - 1 a. no solution b. (-1/2, 5), (2, -5/2) c. (-1/2, -5/2), (2,5)*** d. (1/2, 5/2), (2, 5) 2.how many real

  3. Math

    Solve the following system of equations by graphing. If the system is inconsistent or the equations are dependent, say so 5x-2y=3 10x-4y=6 I know the equations are dependent but I am confused on what is the solution set I'm

  4. Algebra

    Josephine solved a quadratic equation: (x+6)^2=49. Her work is shown below. Step 1: √(x+6)^2 = √49 Step 2: x+6 = 7 Step 3: x = 7−6 Step 4: x = 1 In which step did Josephine make an error? A- step 4 B- step 1 C- step 2 D-

  1. Algebra2

    Find the polynomials roots to each of the following problems: #1) x^2+3x+1 #2) x^2+4x+3=0 #3) -2x^2+4x-5 #3 is not an equation. Dod you omit "= 0" at the end? #2 can be factored into (x+1)(x+3) = 0, so the roots are x=-1 and -3.

  2. math

    Consider the sequence of steps to solve the equation: 5(x - 3) = 7x 2 Step 1 ⇒ 10(x - 3) = 7x Step 2 ⇒ 10x - 30 = 7x Step 3 ⇒ 3x - 30 = 0 Step 4 ⇒ 3x = 30 Step 5 ⇒ x = 10 Identify the property of equality which yields

  3. Pre-Calculus

    How do you solve this quadratic function? f(x) = −x2 + 10x

  4. algebra 1

    Use elimination to solve the the system of equations given by 10x-4y=-20 and 10x-5y=35. I can't figure this one out. Whenever I check my work, it is incorrect.

You can view more similar questions or ask a new question.