how to complete the square for
-2x^2+3x-1?
can you plz help me, and any help is greatly appricieted.
thnx a lot for your effort.
step1
always factor out the coefficient of the x^2 term, unless it is already 1
this guarantees that it will be just x^2
-2x^2+3x-1
= -2[x^2 - 3/2 + ...] - 1
now take 1/2 of the coefficient of the x term, square that result, and then add and subtract it
1/2 of -3/2 is -3/4, which when squared is 9/16
= -2[x^2 - 3/2 + 9/16 - 9/16] - 1
your first 3 terms inside the bracket are your "perfect square"
= -2[x - 3/4)^2 - 9/16] - 1
now multiply through by the -2
= -2(x - 3/4)^2 + 9/8 - 1
simplify the end part
= -2(x - 3/4)^2 + 9/8 - 8/8
= -2(x - 3/4)^2 + 1/8
the middle term should of course be (-3/2)x in two of the above lines.
I am pretty sure you know where the typo error is.
r u there gagan
To complete the square for the quadratic expression -2x^2 + 3x - 1, you can follow these steps:
Step 1: Put the quadratic expression in the form ax^2 + bx + c.
The given expression is already in this form, so we can proceed to the next step.
Step 2: Take the coefficient of x (in this case, 3) and divide it by 2. Square the result.
(3 รท 2)^2 = (3/2)^2 = 9/4
Step 3: Add the result from Step 2 to both sides of the equation.
-2x^2 + 3x - 1 + 9/4 = -2x^2 + 3x + 7/4
Step 4: Rewrite the first three terms as a perfect square trinomial.
The first three terms can be rewritten as (-2x^2 + 3x + 9/4) = (-2(x^2 - (3/2)x + 9/8))
Step 5: Factor the perfect square trinomial from Step 4.
(x^2 - (3/2)x + 9/8) can be factored as (x - 3/4)^2
Step 6: Simplify the right side of the equation.
-2(x - 3/4)^2 + 7/4 = (-2(x - 3/4)^2 + 7)/4
Therefore, the expression -2x^2 + 3x - 1 can be completed by rewriting it as -2(x - 3/4)^2 + 7/4.
By completing the square, we have reexpressed the quadratic equation in vertex form, where the vertex is at (3/4, 7/4).