Restate the question: How do you complete the square when the coefficient of the x-term is odd?
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Compare x2 +bx+c with (x+a)2 = x2 +2ax +a2: b has to equal 2a, which gives a = b/2. . . . For example: Add a constant term to x2 +7x to make a perfect square. b=7, so a=7/2. You have to add (7/2)2 =72/22 = 49/4. . . . x2 + 7x + 49/4 = (x+7/2)2.
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Write it in the form ax2 + bx + c. It is a perfect square if b2 = 4ac
the degree of trinomial is the sum of the variables exponents
You ignore the constant part (-3 in this case), calculate half of the coefficient of the linear part (8, in this case), and square this half. (1/2 of 8 is 4; the square of 4 is 16). This gives you the perfect square x2 + 8x + 16. To get something equivalent to the original expression, you must both add and subtract 16, and include the term which I previously ignored (-3): x2 + 8x + 16 - 16 - 3, which you can write as (x2 + 8x + 16) - 16 - 3. The part within parentheses is the perfect square.
Multiplying a number by its self is called squaring a number When using exponents, the second power of a whole number is called a perfect square too, For example, write 4 squared = 16, 7 squared = 49, and 265 = 70, 225 are all perfect squares.
(2x + 5)(6x - 5)