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Siyavula textbooks: grade 10 Equations and inequalities Solving quadratic equations
You can combine these in many ways and so the best way to develop your intuition for the best thing to do is practice problems. A combined set of operations could be, for example,
1
a
x
2
+
b
x
=
c
(
1
a
x
2
+
b
x
)
-
1
=
(
c
)
-
1
(
invert both sides
)
a
x
2
+
b
x
1
=
1
c
a
x
2
+
b
x
=
1
c
(
a
x
2
+
b
x
)
2
=
(
1
c
)
2
(
square both sides
)
a
x
2
+
b
x
=
1
c
2
Square both sides of the equation
Both sides of the equation should be squared to remove the square root sign.
x
+
2
=
x
2
Write equation in the form
a
x
2
+
b
x
+
c
=
0
x
+
2
=
x
2
(
subtract
x
2
from both sides
)
x
+
2
-
x
2
=
0
(
divide both sides by
-
1
)
-
x
-
2
+
x
2
=
0
x
2
-
x
+
2
=
0
Factorise the quadratic
x
2
-
x
+
2
The factors of
x
2
-
x
+
2 are
(
x
-
2
)
(
x
+
1
) .
Write the equation with the factors
(
x
-
2
)
(
x
+
1
)
=
0
Determine the two solutions
We have
x
+
1
=
0
or
x
-
2
=
0
Therefore,
x
=
-
1 or
x
=
2 .
Check whether solutions are valid
Substitute
x
=
-
1
x
+
2
=
x :
LHS
=
(
-
1
)
+
2
=
1
=
1
but
RHS
=
(
-
1
)
Therefore LHS
≠RHS. The sides of an equation must always balance, a potential solution that does not balance the equation is not valid. In this case the equation does not balance.
Therefore
x
â‰
-
1 .
Now substitute
x
=
2 into original equation
x
+
2
=
x :
LHS
=
2
+
2
=
4
=
2
and
RHS
=
2
Therefore LHS = RHS
Therefore
x
=
2 is the only valid solution
Write the final answer
x
+
2
=
x for
x
=
2 only.
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Solve the equation:
x
2
+
3
x
-
4
=
0 .
Check if the equation is in the form
a
x
2
+
b
x
+
c
=
0
The equation is in the required form, with
a
=
1 .
Factorise the quadratic
You need the factors of 1 and 4 so that the middle term is
+
3 So the factors are:
(
x
-
1
)
(
x
+
4
)
Solve the quadratic equation
x
2
+
3
x
-
4
=
(
x
-
1
)
(
x
+
4
)
=
0
Therefore
x
=
1 or
x
=
-
4 .
Check the solutions
1
2
+
3
(
1
)
-
4
=
0
(
-
4
)
2
+
3
(
-
4
)
-
4
=
0
Both solutions are valid.
Write the final solution
Therefore the solutions are
x
=
1 or
x
=
-
4 .
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Find the roots of the quadratic
equation
0
=
-
2
x
2
+
4
x
-
2 .
Determine whether the equation is in the form
a
x
2
+
b
x
+
c
=
0 , with no
common factors.
There is a common factor: -2.
Therefore, divide both sides of the equation by -2.
-
2
x
2
+
4
x
-
2
=
0
x
2
-
2
x
+
1
=
0
Factorise
x
2
-
2
x
+
1
The middle term is negative. Therefore, the factors are
(
x
-
1
)
(
x
-
1
)
If we multiply out
(
x
-
1
)
(
x
-
1
) , we get
x
2
-
2
x
+
1 .
Solve the quadratic equation
x
2
-
2
x
+
1
=
(
x
-
1
)
(
x
-
1
)
=
0
In this case, the quadratic is a perfect square, so there is only one solution
for
x :
x
=
1 .
Check the solution
-
2
(
1
)
2
+
4
(
1
)
-
2
=
0 .
Write the final solution
The root of
0
=
-
2
x
2
+
4
x
-
2 is
x
=
1 .
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Solving quadratic equations
Solve for
x :
(
3
x
+
2
)
(
3
x
-
4
)
=
0
Solve for
x :
(
5
x
-
9
)
(
x
+
6
)
=
0
Solve for
x :
(
2
x
+
3
)
(
2
x
-
3
)
=
0
Solve for
x :
(
2
x
+
1
)
(
2
x
-
9
)
=
0
Solve for
x :
(
2
x
-
3
)
(
2
x
-
3
)
=
0
Solve for
x :
20
x
+
25
x
2
=
0
Solve for
x :
4
x
2
-
17
x
-
77
=
0
Solve for
x :
2
x
2
-
5
x
-
12
=
0
Solve for
x :
-
75
x
2
+
290
x
-
240
=
0
Solve for
x :
2
x
=
1
3
x
2
-
3
x
+
14
2
3
Solve for
x :
x
2
-
4
x
=
-
4
Solve for
x :
-
x
2
+
4
x
-
6
=
4
x
2
-
5
x
+
3
Solve for
x :
x
2
=
3
x
Solve for
x :
3
x
2
+
10
x
-
25
=
0
Solve for
x :
x
2
-
x
+
3
Solve for
x :
x
2
-
4
x
+
4
=
0
Solve for
x :
x
2
-
6
x
=
7
Solve for
x :
14
x
2
+
5
x
=
6
Solve for
x :
2
x
2
-
2
x
=
12
Solve for
x :
3
x
2
+
2
y
-
6
=
x
2
-
x
+
2
Source:
OpenStax, Siyavula textbooks: grade 10 maths [caps]. OpenStax CNX. Aug 03, 2011 Download for free at http://cnx.org/content/col11306/1.4
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