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Answers
Step-by-step explanation:
Completing the square for quadratic equations in which the quadratic coefficient is one is relatively straightforward. Things get slightly more complicated when the quadratic coefficient is greater than one, but the method is essentially the same, with the addition of an additional step. Consider the following quadratic equation:
5x 2 - 4x - 2 = 0
The additional step required here is to divide both sides of the equation by the quadratic coefficient, which in this case is five:
x 2 - 4 x - 2 = 0
5 5
Since five is not a factor of either four or two, we will write the linear coefficient and constant terms as rational numbers (fractions). As before, move the constant term to the right hand side of the equation:
x 2 - 4 x = 2
5 5
We can now write the left hand side of the equation as a complete square, and add the a 2 term to the right-hand side of the equation:
( x - 2 ) 2 = 2 + 4 ⇒ ( x - 2 ) 2 = 14
5 5 25 5 25
Now take the square root of each side:
x - 2 = ± √ 14 ⇒ x = 2 ± √ 14
5 25 5 25
You could of course have rewritten the quadratic equation (as a monomial) as:
x 2 - 0.8x - 0.4 = 0
This leads to:
(x - 0.4) 2 = 0.4 + 0.16 ⇒ (x - 0.4) 2 = 0.56 ⇒ x - 0.4 = √0.56
From which we get the following values of x:
x ≈ 0.4 ± 0.75 ⇒ x ≈ 1.15 or x ≈ -0.35
Note that taking the square root of 0.56 does not give exactly 0.75. We have rounded the answer to two decimal places. Whether this is accurate enough depends on the nature of the overall problem you are trying to solve. In an exam scenario, you should give answers in the form requested, when this is specified. The procedure for solving quadratic equations in the form ax 2 + bx + c = 0 by completing the square can be summarised as follows:
divide both sides of the equation by the quadratic coefficient
move the constant term to the right-hand side of the equation (this is the same as subtracting the constant term from both sides)
obtain a value for a (this will be the linear coefficient divided by two)
rewrite the left hand side of the equation as (x + a) 2, substituting the actual value of a that you calculated in the previous step
add the value of a 2 to the right-hand side of the equation
take the square root of both sides of the equation
subtract a from both sides of the equation
This will leave you with an equation in the form:
x = -a ± √(constant term + a 2)
The algebraic relationship between the value of a in the above equation and the quadratic and linear coefficients in the generalised form of the quadratic equation is very important, as it can be used to derive the quadratic formula. The quadratic formula, its derivation, and its usage will be discussed on another page
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