Download If we have the equation x2 = 16, what are the solutions to the

If we have the equation x2 = 16, what are the solutions to the equation?
Since the square of a positive or negative number are always positive,
this equation has two solutions, namely x = −4 or x = 4.
What about the equation x2 = 10? 10 is not a perfect square, but
using radicals, we can come up with two possible solutions too:
√
√
x = − 10 or x = 10
In general, all equations of the form x2 = a, where a is a positive
√
√
number, will always give two solutions, namely x = − a or x = a.
We use the symbo ± to mean “plus or minus”. So we will write x =
√
± a.
E.g. Solve the equation: x2 = 7
√
Ans: x = ± 7
E.g. Solve the equation: x2 = 35
√
Ans: x = ± 35
E.g. Solve the equation: x2 = 300
√
√
√
Ans: x = ± 300 = ± 100 · 3 = ±10 3
Notice that if x2 = b and b is a negative number, then we have no (real
number) solution, since the square of a real number is always positive.
Example: Solve: x2 = −5
Ans: The equation has no (real number) solution.
In using the property of square root to solve an equation of the form
x2 = a, the expression being squared does not have to be just x, but
can be more complicated form:
E.g. Solve the equation: (x − 3)2 = 7
Ans: (x−3) is the expression being squared (it is our new x), following
a similar example, we have:
(x − 3)2 = 7
1
√
x−3=± 7
√
x=3± 7
E.g. Solve the equation: (2x + 5)2 = 23
Ans:
(2x + 5)2 = 23
√
2x + 5 = ± 23
√
2x = −5 ± 23
√
−5 ± 23
x=
2
E.g. Solve the equation: (4x + 7)2 = 8
Ans:
(4x + 7)2 = 8
√
√
4x + 7 = ± 8 = ±2 2
√
4x = −7 ± 2 2
√
−7 ± 2 2
x=
4
Completing the Square
Some quadratic equations cannot be solved by factoring. For example,
x2 −8x+1 = 0 cannot be solved by factoring because the left hand side
cannot be factored. Is there another way to try to solve this equation?
We know from the property of radicals that if an equation is of the
√
form x2 = a, where a is a positive number, then x = ± a are the two
solutions to the equation. Is there a way to turn x2 − 8x + 3 into a
form that is a perfect square?
If we look at (x + b)2 = (x + b)(x + b) = x2 + 2bx + b2 , we see that
a quadratic expression is a perfect square if the constant term (the b2
term) is half of the middle term (the 2b term) squared. This gives us
a way to turn a quadratic expression into a perfect square:
Make sure that the coefficient on x2 is 1. Take half of the coefficient
2
of the x term, square the result, and use that as the constant term.
E.g.
x2 − 8x + 3 = 0
We first move the constant of 1 to the right side, since we need a new
constant term:
x2 − 8x = −3
We take half of the coefficient of x, that is, we take half of −8, we get
−4.
We square this number, (−4)2 = 16. This is our new constant term.
We add this term to both sides of the equation:
x2 − 8x + 16 = −3 + 16
x2 − 8x + 16 = 13
The left hand side is now a perfect square, we can factor it:
(x − 4)2 = 13
The property of radical/squares tells us that
√
x − 4 = ± 13
Solving for x gives:
√
x = 4 ± 13
In other words, we get two solutions to the given equation, they are:
√
√
x = 4 + 13 or x = 4 − 13
This method is called The Method of Completing the Square.
What if the coefficient of x2 is not 1? We would then divide both sides
of the equation by the appropriate constant to make the coefficient of
x2 to be 1 first:
E.g.
3x2 + x − 3 = 0
For this equation, the coefficient of x2 is 3, not 1. In order to apply the
method of completing the square, we divide both sides of the equation
3
by 3 first:
3x2 + x − 3 0
=
3
3
1
x2 + x − 1 = 0
3
We now apply the same method again. Note that this time, fractions
will be involved.
1
x2 + x = 1
3
1
1
1
We take half of , which is , and square this number, which gives .
3
6
36
This is the new constant we want to use:
1
1
1
x2 + x +
=1+
3
36
36
1
1
37
x2 + x +
=
3
36 36
Factor the left hand side gives:
1
x+
6
!2
=
37
36
v
u
u 37
1
x + = ±t
6
36
√
37
1
x+ =±
6
6
√
1
37
x=− ±
6
6
Unlike the factoring method, the method of completing the square
works on any quadratic equations, and will always give us any/all
solution to a quadratic equation with real coefficients.
Quadratic Formula:
The method of completing the square can be used to solve any quadratic
equation with real coefficient. The only problem with it, though, is
that it is rather tedious to apply. Instead, we apply the method of
completing the square once on a generic quadratic equation to come
up with a formula:
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General quadratic equation:
ax2 + bx + c = 0
We first divide both sides of the equation by a to make the coefficient
of x2 to be 1.
ax2 + bx + c 0
=
a
a
b
c
x2 + x + = 0
a
a
b
c
x2 + x = −
a
a
b
b
Taking half of gives , and we add the square of this to both sides
a
2a
of the equation, this gives:
b 2
c
b 2
b
=− +
x + x+
a
2a
a
2a
The left hand side is now a perfect square and can be factored:
!
2
b
x+
2a
!2
b
x+
=
2a
c
b
=− +
a
2a
!
!2
v
u
u
u c
±t−
!2
v
u
u
u c
t−
!2
b
x=− ±
2a
b
+
a
2a
b
+
a
2a
Simplify the right hand side gives us the Quadratic Formula:
x=
−b ±
√
b2 − 4ac
2a
The quadrtic formula is a wrapped up version of the method of completing the square. It can be used to solve any quadratic equation just
like the method of completing square. It is, however, more efficient
than completing the square since it only gives the answer.
In the quadratic formula, the expression inside the radical, b2 − 4ac,
is called the discriminent, and its value determines the number of real
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number solutions the quadratic equation has:
If b2 − 4ac > 0, the quadratic equation has 2 real number solutions.
If b2 − 4ac = 0, the quadratic equation has 1 real number solution.
If b2 − 4ac < 0, the quadratic equation has no real number solution.
In applying the quadratic formula, make sure that you recognize what
are the coefficients a, b, and c.
E.g.
x2 + 3x − 5 = 0
In this equation, a = 1, b = 3, c = −5
The solution is
q
√
−3 ± (3)2 − 4(1)(−5) −3 ± 29
x=
=
2(1)
2
E.g.
−2x2 − 2x + 1 = 0
In this equation, a = −2, b = −2, c = 1
The solution is
q
√
−(−2) ± (−2)2 − 4(−2)(1) 2 ± 12
1 √
=
=− ± 3
x=
2(−2)
−4
2
E.g.
3x2 − 4x − 5 = 0
In this equation, a = 3, b = −4, c = −5
The solution is
q
√
√
−(−4) ± (−4)2 − 4(3)(−5) 4 ± 76 2
19
x=
=
= ±
2(3)
6
3
3
E.g.
x2 − x + 13 = 10
We cannot try to use the quadratic formula on this equation at the
moment. Instead, make the right hand side 0 first:
x2 − x + 3 = 0
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Now, a = 1, b = −1, c = 3
The solution is
q
√
−(−1) ± (−1)2 − 4(1)(3) 1 ± −11
x=
=
2
2
Note that the number inside the radical is a negative number, −11.
Since the radical of a negative number is not a real number, we say
that the above equation have no real number solution.
While the quadratic formula (or completing square) can be used to
solve any quadratic equation, in the case that an expression can be
factored, it is still much faster to use the factoring method. Therefore,
in trying to solve a quadratic equation, it is still a good idea to try
to solve by factoring first. Use quadratic formula only if you cannot
factor the expression.
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