Quadratic Equation
This is a guide on quadratic equations.
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We know how to solve equations like:
\(3x - 2 = 0\)
In such an equation, x appears only in the first power. We shall now consider
the next most difficult case, when x appears to the second power.
Consider the equation:
\(x^{2} - 3x + 1 = 0\)
We wish to solve for x, that is, determine all values for x which satisfy this
equation. We shall ultimately derive a general formula for this. Before deriving
the formula, we carry out on this special example the method used to derive the
general formula. Solving our equation amounts to solving:
\(x^{2} - 3x = -1\)
We wish to add a number to both sides of this equation so that the left-hand
side becomes a square, of the form:
\((x - s)^{2}\)
We know that:
\((x - s)^{2} = x^{2} - 2sx + s^{2}\)
So, we need 2s = 3 or s = 3/2. Adding:
\(\frac{3}{2} ^{2} \)
to each side of the equation, we find:
\(x^{2} - 3x + \frac{9}{4} = -1 + \frac{9}{4} = \frac{5}{4}\)
The left hand side has been adjusted so that it is a square:
\(x^{2} - 3x + \frac{9}{4} = (3 - \frac{3}{2})^{2}\)
Solving this equation amounts to solving:
\((x - \frac{3}{2})^{2} = \frac{5}{4}\)
We can now take the square root, and we find that x is a solution if and only
if:
\(x - \frac{3}{2} = \pm \sqrt{5}{4}\)
There are two values here:
\(x = \frac{3}{2} + \sqrt{5}{4}\)
and
\(x = \frac{3}{2} - \sqrt{5}{4}\)
We wish to solve the equation
\(x^{2} + 2x + 2 = 0\)
We apply the same method as before. We will be solving
\(x^{2} + 2x = -2\)
We add 1 to both sides, so that we are able to express the left hand in the form
\(x^{2} + 2x + 1 = (x + 1)^{2}\)
It is equivalent to solving
\((x + 1)^{2} = -2 + 1 = -1\)
A negative number cannot possibly be the square of a real number, and we
conclude that our equation does not have a solution in real numbers.
We wish to solve the equation
\(2x^{2} - 3x - 5 = 0\)
This is solving
\(2x^{2} - 3x = 5\)
This time, we see that x is multiplied by 2. To reduce our problem to one similar
to those already considered, we divide the whole equation by 2, and solving is
equivalent to
\(x^{2} - \frac{3}{2}x = \frac{5}{2}\)
We can now complete the square on the left as we did before. We need to find a
number s such that
\(x^{2} - \frac{3}{2}x = x^{2} - 2sx\)
This means that s = 3/4. Adding s squared to both sides we find
\(x^{2} - \frac{3}{2}x + \frac{9}{16} = \frac{5}{2} + \frac{9}{16} =
\frac{49}{16}\)
Expressing the left hand side as a square is equivalent to
\((x - \frac{3}{4})^{2} = \frac{49}{16}\)
We can now solve for x
\(x - \frac{3}{4} = \pm \sqrt{49}{16}\)
or
\(x = \frac{3}{4} \pm \sqrt{49}{16}\)
This answer is correct but we should always watch for simplifications
\(\sqrt{49}{16} = \frac{7}{4}\)
So we then have
\(x = \frac{3}{4} \pm \frac{7}{4}\)
Which gives
\(x = \frac{10}{4}\)
and
\(x = \frac{-4}{4} = -1\)
If a,b,c are real numbers and a does not equal 0, the solutions of the equation
\(ax^{2} + bx + c = 0\)
are given by the formula
\(x = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a}\)
If the square root erm is negative then the equation has no solution in real
numbers
Solving our equation amounts to solving
\(ax^{2} + bx = -c\)
Dividing by a, we see that it is equivalent to solving
\(x^{2} + \frac{b}{a}x = - \frac{c}{a}\)
To complete the square on the left, we need
\(x^{2} + \frac{b}{a}x = x^{2} + 2sx\)
So s = b/2a. We then add
\(s^{2} = \frac{b^{2}}{4a^{2}}\)
to both sides and find the equivalent equation
\((x + \frac{b}{2a})^{2} = -\frac{c}{a} + \frac{b^{2}}{4a^{2}}\)
\(= \frac{b^{2} - 4ac}{4a^{2}}\)
If
\(b^{2} - 4ac\)
is negative, then
\(\frac{b^{2} - 4ac}{4a^{2}}\)
is negative and so cannot be the square of a real number. So the equation has no
real solution.
If
\(b^{2} - 4ac > 0\)
then we get precisely one solution for the quadratic equation.
\(x = \frac{-b}{2a}\)
If
\(b^{2} - 4ac > 0\)
then we get two solutions, one positive and one negative.
The quadratic formula is so important that it should be memorized
Solve the equation
\(3x^{2} - 2x + 1 = 0\)
\(x = \frac{-(-2) \pm \sqrt{(-2)^{2} - 4*3}}{2*3}\)
\(x = \frac{4 \pm \sqrt{-8}}{6}\)
In this case we see that the expression
\(b^{2} - 4ac\)
under the square root sign is negative and so our equation has no real solution
in the real numbers.
Solve the equation
\(2x^{2} + 3x - 4 = 0\)
Use the formula to get
\(x = \frac{-3 \pm \sqrt{9 - 4*2*(-4)}}{2*2}\)
\(x = \frac{-3 \pm \sqrt{9 + 32}}{4}\)
\(x = \frac{-3 \pm \sqrt{41}}{4}\)
This is our answer and we get a positive and negative answer of the above
answer.
In the proof of our theorem concerning the solutions of the quadratic equation,
we needed to operate with addition, multiplication, and square roots. If we knew
that the real numbers could be extended to a larger system of numbers in which
these operations were valid, including the possibility of taking square roots of
negative real numbers, then our formula would be valid in this bigger system of
numbers, and would again give the solutions of the equation in all cases.