how do i solve this completing the square problem: 3x^2+12x+10=0
divide by 3
x^2+4x+10/3=0
x^2+4x + ???= -10/3 + ???
well, look at the second term. take one half of it, and square it, add to both sides.
x^2+4x+4=-10/3+4= 10/3+12/3
(x+2)^2= 2/3
take the square root of each side.
x+2= +-sqrt 2/3
x=-2+-sqrt 2/3
check that.
To solve the given quadratic equation, 3x^2 + 12x + 10 = 0, we can use the method of completing the square. Here's how you can approach it:
Step 1: Make sure the coefficient of x^2 is 1. In this case, it already is, so we can proceed directly to the completing the square process.
Step 2: Move the constant term (the term without an x) to the other side of the equation. Subtract 10 from both sides:
3x^2 + 12x = -10
Step 3: Divide the coefficient of x by 2, square the result, and add it to both sides of the equation. The result should create a perfect square trinomial on the left side. In this case, the coefficient of x is 12, so we divide it by 2 to get 6. Squaring 6 gives us 36, so we add 36 to both sides:
3x^2 + 12x + 36 = -10 + 36
3x^2 + 12x + 36 = 26
Step 4: Factor the perfect square trinomial on the left side of the equation. To factor it, take the square root of the first term, the square root of the last term, and rewrite the middle term as the sum of these square roots:
(√3x + 6)^2 = 26
Step 5: Take the square root of both sides of the equation, considering both the positive and negative square roots:
√(√3x + 6)^2 = ±√26
√3x + 6 = ±√26
Step 6: Isolate x by subtracting 6 from both sides of the equation:
√3x = ±√26 - 6
Step 7: Square both sides of the equation to eliminate the square root:
(√3x)^2 = (±√26 - 6)^2
3x = (±√26 - 6)^2
Step 8: Simplify and solve for x by dividing both sides of the equation by 3:
x = [(±√26 - 6)^2] / 3
Therefore, the solutions to the quadratic equation 3x^2 + 12x + 10 = 0 are:
x = [(√26 - 6)^2] / 3 and x = [(-√26 - 6)^2] / 3
3x^2+12x+10=0
3(x^2+4x+10/3)=0
(x+2)²-4+10/3=0
(x+2)²-(√(2/3))²=0
(x+2+√(2/3))(x+2-√(2/3))=0
therefore
x=-2-√(2/3) or -2+√(2/3)