Download A6 Quadratic equations

Solving by factorization
1. Solving by factorization
2. Solving by completing the square
3. Solving by using the quadratic equation formula
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Solving by factorization
We can solve the quadratic equation x2 + 4x – 45 = 0 in full
by factorizing the expression on the left-hand side.
This means that we can write the equation in the form
(x + ….)(x + ….) = 0
We need to find two integers that add together to make 4 and
multiply together to make –45.
Because –45 is negative, one of the numbers must be
positive and one must be negative.
By considering the factors of 45 we find that the two numbers
must be 9 and –5.
We can therefore write x2 + 4x – 45 = 0 as
(x + 9)(x – 5) = 0
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Solving by factorization
When two numbers multiply together to make 0, one of the
numbers must be 0, so if
(x + 9)(x – 5) = 0
we can conclude that either
x+9=0
or
x–5=0
This gives us two solutions that solve the quadratic equation:
x=–9
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and
x=5
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Solving by factorization – Example 1
Solve the equation x2 = 3x by factorization.
Start by rearranging the equation so that the terms are on the
left-hand side,
x2 – 3x = 0
Factorizing the left-hand side gives us
x(x – 3) = 0
So
x=0
or
x–3=0
x=3
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Solving by factorization – Example 2
Solve the equation x2 – 5x = –4 by factorization.
Start by rearranging the equation so that the terms are on the
left-hand side.
x2 – 5x + 4 = 0
We need to find two integers that add together to make –5
and multiply together to make 4.
Because 4 is positive and –5 is negative, both the integers
must be negative. These are –1 and –4.
Factorizing the left-hand side gives us
(x – 1)(x – 4) = 0
x – 1 = 0 or
x–4=0
x=1
x=4
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Solving by factorization – Practice 1
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Completing the square
Adding 9 to the expression x2 + 6x to make it into a perfect
square is called completing the square.
We can write
x2 + 6x = x2 + 6x + 9 – 9
If we add 9 we then have to subtract 9 so that both sides are
still equal.
By writing x2 + 6x + 9 we have completed the square and so
we can write this as
x2 + 6x = (x + 3)2 – 9
In general,
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x2
+ bx = x +
b 2
2
–
b 2
2
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Completing the square
Complete the square for x2 – 10x.
Compare this expression to (x – 5)2 = x2 – 10x + 25
x2 – 10x = x2 – 10x + 25 – 25
= (x – 5)2 – 25
Complete the square for x2 – 3x.
Compare this expression to (x – 1.5)2 = x2 – 3x + 2.25
x2 – 3x = x2 – 3x + 2.25 – 2.25
= (x + 1.5)2 – 2.25
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Expressions in the form x2 + bx
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Completing the square
How can we complete the square for
x2 + 8x + 9?
Look at the coefficient of x.
This is 8 so compare the expression to (x + 4)2 = x2 + 8x + 16.
x2 + 8x + 9 = x2 + 8x + 16 – 16 + 9
= (x + 4)2 – 7
In general,
x2
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+ bx + c = x +
b 2
2
–
b
2
2
+c
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Completing the square
Complete the square for x2 + 12x – 5.
Compare this expression to (x + 6)2 = x2 + 12x + 36
x2 + 12x – 5 = x2 + 12x + 36 – 36 – 5
= (x2 + 6) – 41
Complete the square for x2 – 5x + 16
Compare this expression to (x – 2.5)2 = x2 – 5x + 6.25
x2 – 5x + 16 = x2 – 5x + 6.25 – 6.25 + 16
= (x2 – 2.5) + 9.75
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Expressions in the form x2 + bx + c
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Completing the square
When the coefficient of x2 is not 1, quadratic equations in the
form ax2 + bx + c can be rewritten in the form a(x + p)2 + q by
completing the square.
Complete the square for 2x2 + 8x + 3.
Start by factorizing the first two terms by dividing by 2,
2x2 + 8x + 3 = 2(x2 + 4x) + 3
By completing the square, x2 + 4x = (x + 2)2 – 4 so,
2x2 + 8x + 3 = 2((x + 2)2 – 4) + 3
= 2(x + 2)2 – 8 + 3
= 2(x + 2)2 – 5
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Completing the square
Complete the square for 5 + 6x – 3x2.
Start by factorizing the the terms containing x’s by –3.
5 + 6x – 3x2 = 5 – 3(–2x + x2)
5 + 6x – 3x2 = 5 – 3(x2 – 2x)
By completing the square, x2 – 2x = (x – 1)2 – 1 so,
5 + 6x – 3x2 = 5 – 3((x – 1)2 – 1)
= 5 – 3(x – 1)2 + 3
= 8 – 3(x – 1)2
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Expressions in the form ax2 + bx + c
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Solving by completing the square – Example 1
Quadratic equations that cannot be solved by factorization can
be solved by completing the square.
For example, the quadratic equation,
x2 – 4x – 3 = 0
can be solved by completing the square as follows,
(x – 2)2 – 4 – 3 = 0
(x – 2)2 – 7 = 0
simplify:
add 7 to both sides:
(x – 2)2 = 7
square root both sides:
x = 2 + √7
x = 4.646
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x – 2 = ±√7
or
x = 2 – √7
x = –0.646 (to 3 d.p.)
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Solving by completing the square – Example 2
Solve the equation x2 + 8x + 5 = 0 by completing the
square. Write the answer to 3 decimal places.
x2 + 8x + 5 = 0
Completing the square on the left-hand side,
(x + 4)2 – 16 + 5 = 0
(x + 4)2 – 11 = 0
simplify:
add 11 to both sides:
(x + 4)2 = 11
square root both sides:
x + 4 = ±√11
x = –4 + √11
x = –0.683
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or
x = –4 – √11
x = –7.317 (to 3 d.p.)
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Solving by completing the square – Example 3
Solve the equation 2x2 – 4x + 1 = 0 by completing
the square. Write the answer to 3 decimal places.
We can complete the square for 2x2 – 4x + 1 by first
factorizing the terms containing x’s by the coefficient of x2,
2x2 – 4x + 1 = 2(x2 – 2x) + 1
Next complete the square for the expression in the bracket,
= 2((x – 1)2 – 1) + 1
= 2(x – 1)2 – 2 + 1
= 2(x – 1)2 – 1
We can now use this to solve the equation 2x2 – 4x + 1 = 0.
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Solving by completing the square – Example 4
Solve the equation 2x2 – 4x + 1 = 0 by completing
the square. Write the answer to 3 decimal places.
2x2 – 4x + 1 = 0
completing the square:
2(x – 1)2 – 1 = 0
2(x – 1)2 = 1
add 1 to both sides:
(x – 1)2 =
divide both sides by 2:
x–1=±
square root both sides:
x=1+
x = 1.707
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1
2
1
2
or
1
2
x=1–
1
2
x = 0.293 (to 3 d. p)
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Using the quadratic formula
Any quadratic equation of the form,
ax2 + bx + c = 0
can be solved by substituting the values of a, b and c into the
formula,
–b ± b2 – 4ac
x=
2a
This equation can be derived by completing the square on
the general form of the quadratic equation.
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Using the quadratic formula – Example 1
Use the quadratic formula to solve x2 – 7x + 8 = 0.
1x2 – 7x + 8 = 0
–b ± b2 – 4ac
x=
2a
7 ± (–7)2 – (4 × 1 × 8)
x=
2×1
7 ± 49 – 32
x=
2
7 + 17
x=
2
x = 5.562
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or
7 – 17
x=
2
x = 1.438 (to 3 d.p.)
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Using the quadratic formula – Example 2
Use the quadratic formula to solve 2x2 + 5x – 1 = 0.
2x2 + 5x – 1 = 0
–b ± b2 – 4ac
x=
2a
–5 ± 52 – (4 × 2 × –1)
x=
2×2
–5 ± 25 + 8
x=
4
–5 + 33
x=
4
x = 0.186
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or
–5 – 33
x=
4
x = –2.686 (to 3 d.p.)
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Using the quadratic formula – Example 3
Use the quadratic formula to solve 9x2 – 12x + 4 = 0.
9x2 – 12x + 4 = 0
–b ± b2 – 4ac
x=
2a
12 ±  (–12)2 – (4 × 9 × 4)
x=
2×9
12 ± 144 – 144
x=
18
12 ± 0
x=
18
2
There is only one solution, x = 3
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Using the quadratic formula – Example 4
Use the quadratic formula to solve x2 + x + 3 = 0.
1x2 + 1x + 3 = 0
–b ± b2 – 4ac
x=
2a
–1 ± 12 – (4 × 1 × 3)
x=
2×1
–1 ± 1 – 12
x=
2
–1 ± –11
x=
2
We cannot find –11 and so there are no solutions.
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