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COLLEGE ALGEBRA MATH 110 |
Southwestern Adventist University |
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| Distance Education | Lawrence E. Turner, Jr., Ph.D. | |
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Solving Quadratic Equations
ax + b = 0, a ≠ 0
where a and b are given coefficients and x is the unknown value to be determined, is relatively straight forward to solve; that is, find the value of x that satisfies or makes the equation true. We use familiar algebraic manipulations to solve for x.
On the other hand, the general quadratic is not so straight forward to solve:
ax2 + bx + c = 0, a ≠ 0
We do know that since it is a second degree polynomial, there should be two solutions. (There might be a single value that is repeated.) Can every quadratic actually be solved? Can we solve it?
The first answer is yes, although this is not true for polynomials higher than degree five. We can find exact solutions, not just numeric approximations.
The second answer is also an emphatic yes!.
Special Cases
There are two special cases that are easy to solve.First special case: c = 0
The quadratic equation becomes:
ax2 + bx = 0
This can be readily factored into two factors as:
x(ax + b) = 0
Values of x that make this true are those that make each factor zero:
x = 0 and (ax + b) = 0
The second is just a linear equation with a solution:
x = −b/a
Therefore, we get two solutions:
x = 0 and x = −b/a
Second special case: b = 0
The quadratic equation becomes:
ax2 + c = 0
Since there is no x term, this is also readily solved first for x2:
x2 = −c/a = 0
Then by taking the square root of both sides we can obtain x. However, to do this we must recall that there are two values that when squared will produce, say, a 4. Namely +2 and −2. We can write these as ±2. When we take the square root of both sides of the equation we must include both of these possibilities for x. Thus, we get:
x = ±√(−c/a)
Solving where no coefficients are zero
There are three basic methods: factoring, completing the square, and using the quadratic formula; although the last one is derived by completing the square, the thought processes and work in using the quadratic formula make it one of the major "independent" methods.Factoring
If the given quadratic equation can be readily factored, then this is perhaps the quickest way. Written as two factors the quadratic equation looks like:
ax2 + bx + c = (px + r)(qx + s) = 0
The solutions are: x = −r/p and x = −s/q
Clearly, a = pq, c = rs, and b = ps + rq; however, to use these to find p, q, r, and s is not necessarily easy or straight forward. To illustrate the factoring method, it is necessary to use a specific example.
x2 + 5x + 6 = 0
can be factored as:
x2 + 5x + 6 = (x + 2)(x + 3) = 0
with solutions: x = −2 and x = −3.
Completing the Square
This method is motivated by the observation that if we could rearrange the original quadratic equation as:
a(x + r)2 + s = 0
then we readily find the solutions by isolating (x + r)2 on one side as:
(x + r)2 = −s/a
and then taking the square root of both sides just like we did with the special case of b = 0 above:
(x + r) = ±√(−s/a)
Then solving for x we get:
x = −r ±√(−s/a)
Ok, but how do we get quadratic equation in this state?
The answer is that we can add and subtract a constant term (so that the equality to zero does not change) so that the piece we add along with the first two terms makes a perfect square − this is why this method is called completing the square!
We will start by considering the case where a is 1, then fix that up later. This will make things simpler to grasp. This gives us:
x2 + bx + c = 0
Now, if we square something like (x + r) we get:
(x + r)2 = x2 + 2rx + r2
The right hand side of this last equation can be matched with the original equation. Both have an x2 with a coefficient of 1, an x term but with different coefficients, and a constant term. Now all we have to do is match the coefficients of x-term in the original equation; that is, by equating 2r = b or r = b/2. A perfect square will have a constant term of r2 (which will, in terms of the original coefficient, be (b/2)2). So let us add and subtract this as:
(x2 + 2rx + r2) + c − r2 = 0
or
(x + r)2 + c − r2 = 0
The constant term of c − r2 can be moved to the other side of the equation, then we can take the square root, and finally solve for x as we did above.
Keeping in mind that r = b/2, we see that the procedure is compute (b/2)2 then add and subtract this quantity and regroup the pieces into a single perfect square and a constant term.
| Complete the Square For: x2 + bx + c = 0 Add and subtract (b/2)2 to obtain: [x2 + bx + (b/2)2] + c − (b/2)2 = 0 Rewrite as: [x + (b/2)]2 + c − (b/2)2 = 0 |
If a is not 1, then the process is only slightly more complicated. First factor out an a from the first two terms as:
a[x2 + (b/a)x ] + c = 0
Now we take half of the coefficient of the x-term just like before, in this case (b/a), square it and add and subtract it. However, because we are adding it inside the brackets which are multiplied by a we need to multiply it by a when we subtract it!
a[x2 + (b/a)x + (b/2a)2] + c − a(b/2a)2 = 0
The quantity in the brackets is a perfect square. Therefore we get:
a[x + (b/2a)]2 + c − a(b/2a)2 = 0
This looks a lot worse than it does with specific numbers!
The Quadratic Formula
If we utilize the scheme of Completing the Square on the general quadratic equation:
ax2 + bx + c = 0, a ≠ 0
we obtain a most remarkable formula:
This formula can be used to solve any quadratic equation. Some of the methods given earlier such as the special cases and factoring are perhaps faster when they are readily be used, but the quadratic formula is the one and only method you will ever need to solve any quadratic equation! Therefore, memorize this formula!
In using this formula, there are several precautions:
- Make certain the quadratic equation you are given is rearranged to look like the one above with all the terms on one side with zero on the other. If this is not done, it is too easy to miss the proper sign on a number when substituting into the quadratic formula.
- Substitute all values surrounded by parentheses. If you do not, then it is too easy to miss a sign or worse, turn a multiplication into a subtraction.
Remember that the b2 term for b = −4 is (−4)2 = 16 and not −(4)2 = −16.
- If the quantity inside the radical is negative, then change its sign (make it positive) and write the imaginary unit, i, just in front of the radical.
© 1999, 2000, 2001, 2002, 2003, 2004, 2005, 2006, 2007, 2008 by Lawrence Turner