Quadratic EquationsSolving Quadratic Equations
You already know how to solve
Now let’s think about a more complex class of equations which also contain
where x is a variable and a, b, and c are some specific numbers (called coefficients). Both
Like you saw in the introduction, plotting the graph of a quadratic function in a coordinate system gives a curved shape called a Parabola:
Try changing the values of a, b and c, and see how the parabola changes.
To solve a quadratic equation, we have to find the points where
While linear equations always have exactly one solution, we can see from the diagram that quadratic equations can sometimes have and , , or even .
In the following sections we will discover why that is the case, learn several different methods to find all solutions of a quadratic equation.
Level 0: Taking Square Roots
When trying to solve equations, we often use opposites of mathematical operators. For example, addition and subtraction are opposites, and multiplication and
This can help us to solve some simple quadratic equations:
First, we isolate
Now we take square roots of both sides, remembering to add a ±:
Sometimes we have to do a bit more work to isolate
For every value of
x 2 possible values of x : a positive and a negative one. For example, ifx 2 = ${x*x} , we don't know ifx = ${x} orx = ${'–'+x}. In this case, the quadratic equation has two solutions.Square numbers are always positive. This means that there
x that could satisfyx 2 = − 9 . This equation has no solutions.
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Something about exactness and how to express solutions
As an abbreviation, we sometimes write
EXAMPLE 11: Solve
The only possible difficulty here is that students often forget that in algebra most numbers have two square roots.
EXAMPLE 12: Solve
EXAMPLE 13: Solve
EXAMPLE 14: Solve
PRACTICE 15: Solve: a) x^2=121. b) p^2=40. c) y^2+5=14. d) 2x^2=50. e) x^2=−6.
Level 1:
EXAMPLE 17: Solve
Answer: A tad more complicated but still easy. This problem is saying: “Something squared is 100.” So the something must be 10 or −10. That is:
Example
Solve
Answer: We have: y−4=5 or y−4=−5 yielding: y=9 or y=−1.
WARNING!! Many students like to answer questions like these using the ± symbol. But there is a potential danger. Some will then write the following:
Example 2
Solve 4(p+2)2−16=0
Answer: Add 16:
4(p+2)2=16.
Divide though by 4:
(p+2)2=4.
So
p+2=2 or p+2=−2 yielding: p=0 or p=−4.
Example 3
Solve (x+7)2+9=0
Answer: We have (x+7)2=−9. In the system of real numbers, it is impossible for a quantity squared to be negative. This equation has no solution.
Example 4
Solve (x−1)2=5
Answer: We have:
x−1=√5 or x−1=−√5 So x=1+√5 or x=1−√5.
Example 5
Solve (x+3)2=49
Answer: We have:
x+3=7 or x+3=−7 yielding: x=4 or x=−10.
PRACTICE 23: Solve:
a) (x−1)^2=64 b) (p−3)^2=16 c) (y+1)^2−2=23 d) 3(x−900)^2=300 e) (x−√2)^2=5
PRACTICE 24:
a) How many solutions does (x+7)^2=0 possess? What are they? b) How many solutions does (x+7)^2=−2 have? What are they?
Level 2:
EXAMPLE 25: Solve
Answer: If one is extremely clever one might realize that this is a repeat of example 22:
The quantity
To check this, let’s work out
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Yes, we see that (x+3)2 does equal x2+6x+9. So the question: Solve x2+6x+9=49 is indeed really the question: Solve (x+3)2=49 in disguise.
We have: x+3=7 or x+3=−7 x=4 or x=−10.
So the challenge in level 2 is to recognize more complicated expressions as easy level 1 problems in disguise.
Here are some more examples
Example 2
Solve
Answer: Let’s draw the square for
There is an
Because we want the figure to be a perfectly symmetrical square (squares are good for level 1) the “4x ” must come from two symmetrical pieces: 2x and 2x.
This means we must have the numbers 2 and 2 on the sides of the square, and this is consistent with the final portion being 4.
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Thus we see that
This is easy:
x+2=5 or x+2=−5 x=3 or x=−7.
Example 2
Solve x2−10x+25=169.
Answer: Let’s draw the square for x2−10x+25.
There is an x2 piece that must come from x×x . And we need two (symmetrical) pieces that add to =10x.
This means the need the numbers −5 and −5, which is consistent with the final portion of the square being 25.
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Example 3
Solve x2−20x+100=7.
Answer: We have:
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x2−20x+100=7 (x−10)2=7 x−10=7–√ or x−10=−7–√ x=7–√+10 or x=−7–√+10 □
Example 4
Solve x2+25–√x+5=4.
Answer: Don’t be perturbed by the numbers. Just follow things through as before.
Do you see we have this square?
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x2+25–√x+5=4 (x+5–√)2=4 x+5–√=2 or x+5–√=−2 x=2−5–√ or x=−2−5–√ □
PRACTICE 30: Solve:
a) x2+40x+400=100 b) p2−6p+9=9 c) x2−4x+4=1 d) 3x2−18x+27=12 e) x2−22–√x+2=19
PRACTICE 31: Solve x2+2Bx+B2=D2 in terms of B and D.
Level 3
EXAMPLE 32: Solve x2+8x+15=80.
Answer: Let’s apply the technique of level 2 and draw the box.
The x2 piece must come from x×x. And because we want the symmetry of a square, 8x must come for two pieces 4x each:
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This means we must have the numbers 4 and 4 on the sides of the square, giving us 16 for the remaining piece of the squares, WHICH IS INCONSISTENT WITH THE NUMBER 15 in the problem.
It seems the box method is not of help. We have two options:
- Give up and cry.
- Be an adult and make it work!
Let’s follow option 2 making use of a good piece of general advice: If there is something in life you don’t like and are not happy with, change it!
In this problem we would like the number 15 in x2+8x+15 to actually be a 16. So let’s add one and make it 16!
Of course, to keep the equation balanced, if we add 1 to one side of an equation we need to add 1 to the other side as well:
x2+8x+15+1=80+1.
That is, we have:
x2+8x+16=81.
And the box tells us that this is:
(x+4)2=81 and all now falls into place:
x+4=9 or x+4=−9 x=5 or x=−13
Here are a few more examples:
Example 1
Solve x2−6x+11=27.
The box tells us that the x2 and the −6x pieces want the number 9 to accompanying them, not 11:
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Let’s make that happen. Subtract 2 from both sides:
x2−6x+9=25 (x−3)2=25 x−3=5 or x−3=−5 x=8 or x=−2 □
Example 2
Solve x2−10x+3=0.
Answer: The box tells us that the x2 and the −10x pieces want the number 25 to accompanying them, not 3:
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So let’s add 22 to both sides to make that happen!
x2−10x+3=0 x2−10x+25=22 (x−5)2=22 x−5=2–√2 or x−5=−2–√2 x=5+2–√2 or x=5−2–√2
Example 3
Solve x2−6x=55.
Answer: To obtain a perfect square add 9 to both sides.
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x2−6x+9=64 (x−3)2=64 x−3=8 or x−3=−8 x=11 or x=−5
Example 4
Solve w2+90=22w−31.
Answer: Let’s bring all the terms containing a variable to the left side:
w2−22w+90=−31.
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The box tells us to add 31 to each side:
w2−22w+121=0 (w−11)2=0 w−11=0 w=11 □
Now it's your turn! Try to solve these practice problems:
Problems
Solve these equations:
a) x2+12x−5=40 b) z2+2z+3=11 c) x2−40x+300=69 d) 2f2−16f+30=48 e) x2+100x+2=3
Solve x2+3x+1=5. Don’t be afraid of fractions. You can handle them!
Level 4
EXAMPLE 39: Solve x2+3x+1=5.
Answer: Solving this problem does indeed require the use of fractions:
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Adding
x2+3x+94=5+114 (x+32)2=614 (x+32)2=254 x+32=52 or x+32=–52 x=1 or x=−4 However, most people would prefer to not work with fractions.
The problem here is that the middle term, the 3x, has an odd number for a coefficient. So … If we don’t like that, let’s change it!
Perhaps the easiest way to create an even number in the middle is to multiply everything through by 2, so x2+3x+1=5 becomes:
2x2+6x+2=10.
COMMENT: Some students might say it would be simpler to add x to both sides to then obtain x2+4x+1=5+x. But then we have x’s on both sides of the equation, which is annoying.
Okay … Let’s try the box method on 2x2+6x+2=10.
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But we have a problem now with the very first piece of the equation: We need two symmetrical terms that multiply together to make 2x2. Most students would suggest 2x and x, but this right away ruins the square method we’ve been following in levels 1, 2, and 3.
ALTERNATIVE IDEA: Instead of multiplying through by 2, multiply through by 4.
This again makes the middle term even AND solves the problem with the first term. Let’s see why:
Start with: x2+3x+1=5.
Multiply through by 4:
4x2+12x+4=20.
Now apply the box method:
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We see that 4x2 comes from 2x multiplied with 2x, preserving the symmetry.
Notice that the middle term 12x splits into two equal pieces, 6x plus 6x, as planned, which means we need the numbers 3 and 3 on the sides ( 2x times THREE gives 6x).
We also see that the number we need from the box is 9. Adding 5 to both sides of the equation 4x2+12+9=20 yields:
4x2+12x+9=25.
The box shows that 4x2+12x+9 is really (2x+3)2, so we have a level 1 problem:
(2x+3)2=25 2x+3=5 or 2x+3=−5 2x=2 or 2x=−8 x=1 or x=−4 □
IF THE MIDDLE TERM IS ODD, MULTIPLYING THROUGH BY 4 IS A CLEVER IDEA!
Example 4
Solve x2−5x+6=2.
Answer: Let’s multiply through by 4:
4x2−20x+24=8
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Adding 1 to both sides gives:
4x2−20x+25=9 (2x−5)2=9 2x−5=3 or 2x−5=−3 2x=8 or 2x=2 x=4 or x=1 □
Example 4
Solve x2+x=34.
Answer: Let’s multiply through by 4:
4x2+4x=3
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Adding 1 to both sides gives:
4x2+4x+1=4 (2x+1)2=4 2x+1=2 or 2x+1=−2 2x=1 or 2x=−3 x=12 or x=−32 □
Example 4
Solve p2+7p−2=5
Answer: Let’s multiply through by 4:
4p2+28p−8=20
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Adding 57 to both sides gives:
4p2+28p+49=77 (2p+7)2=77 2p+7=7–√7 or 2p+7=−7–√7 2p=7–√7–7 or 2p=−7–√7–7 p=7√7–72 or p=−7√7–72 □
PRACTICE 43: Solve:
a) x2+11x−5=7 b) z2−3z+1=−1 c) x2−x−1=234 d) x2+5x+12=70 e) x2+3=9
Level 5 Quadratics
EXAMPLE 45: Solve 3x2+5x+1=9.
Answer: This is the first example we’ve encountered with a first term more complicated than just x2. We could divided throughout by 3 and solve instead the equation x2+13x+13=3 and use the box method – and it will work (try it?) – but we will be thick in the midst of fractions.
We have in the previous level multiplied through by 4 and have successfully dealt with 4x2 as 2x×2x. This works because 4 is a perfect square.
So in this problem, let’s try making 3x2 into a perfect square by multiplying through by 3.
9x2+15x+3=27.
The first term is 3x×3x but the middle term is 15x, which has an odd coefficient. To avoid fractions, let’s also multiply through by 4.
36x2+60x+12=108.
This has kept the first term a perfect square – we have 36x2=6x×6x – and has made the second term even. It seems we are set to go!
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The box shows we would like the number 25 to appear. Let’s add 13 to both sides:
36x2+60x+25=121 6x+5)2=121 6x+5=11 or 6x+5=−11 6x=6 or 6x=−16 x=1 or x=−83 Success! □
The previous example illustrates …
THE ULTIMATE BOX METHOD To solve an equation of the form:
ax2+bx+c=d i) MULTIPLY THROUGH BY a (to make the first term a perfect square) ii) MULTIPLY THROUGH BY 4 (to avoid fractions) iii) DRAW THE BOX
and off you go!
THE BOX METHOD WILL NEVER LET YOU DOWN!
Example 1
Solve 7x2−x+1=9.
Answer: Let’s multiply through by 7 to make the leading term a square:
49x2−7x+7=63 and through by 4 to make the second term even (and to preserve the square):
196x2−28x+28=252
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Subtract 27 from each side and we’re good to go!
196x2−28x+1=225 (14x−1)2=225 14x−1=15 or 14x−1=−15 14x=16 or 14x=−14 x=87 or x=−1 □
Example 2
Solve 2x2+3x−3=5.
Answer: Let’s multiply through by 2 to make the first term a perfect square:
4x2+6x−6=10.
Let’s multiply through by 4 and we’ll see the fractions are cleared away:
16x2+24x−24=40.
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Let’s add 33 to each side:
16x2+24x+9=73 (4x+3)2=73 4x+3=73−−√ or 4x+3=−73−−√ 4x=−3+73−−√ or 4x=−3−73−−√ x=−3+73√4 or x=−3−73√4 The numbers weren’t pretty, but the method is straightforward. □
Example 2
Solve −2x2+3x+7=1.
Answer: Let’s multiply through by −2 and then by 4. That is, let’s multiply through by −8.
16x2−24x−56=−8
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16x2−24x+9=−8+9+56 (4x−3)2=57 4x−3=±5–√7 x=3±5√74 Done! □ (QUESTION: Was it dangerous to use the ± symbol here?)
Example 2
Solve 11x2−x+5=0
Answer: Let’s multiply through by 11and by 4, that is, through by 44:
484x2−44x+220=0.
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Subtracting 219 from both sides gives:
484x2−44x+1=−219 (22x−1)2=−219 But there is no number whose square is negative!
The box method is telling us there is no solution to this equation! □
COMMENT: Every example thus far has been crafted to have a solution, but this need not always be the case. For example:
x2=9 has exactly two solutions.
x2=0 has exactly one solution.
x2=−9 has no solutions.
The box method turns every quadratic into an equation of the form:
(something)2=A.
If A is positive, there will be two solutions; if A is zero, there will be one solution; and if A is negative, there will be no solutions.
Problems
PRACTICE 50: Solve:
a) 3x2+7x+5=1 b) 5x2−x−18=0 c) 3x2+x−2=2 d) 2x2−3x=5 e) 10x2−10x=1 f) 2x2−3x+2=0 g) 2x2=9 h) 4−3x2=2−x
a) A rectangle is twice as long as it is wide. Its area is 30 square inches. What are the length and width of the rectangle?
b) A rectangle is four inches longer than it is wide. Its area is 30 square inches. What are the length and width of the rectangle?
c) A rectangle is five inches longer than its width. Its area is 40 square inches. What are the dimensions of the rectangle?
PRACTICE 54: Solve the following quadratic equations:
a) v2−2v+3=27 b) z2+4z=7 c) w2−6w+5=0 d) α2−α+1=74 Also note that x=(x−−√)2. Solve the following disguised quadratics.
e) x−6x−−√+8=0 f) x−2x−−√=−1 g) x+2x−−√−5=10 WATCH OUT! Explain why only one answer is valid for g).
h) 3β–2β−−√=7 i) 2u4+8u2−5=0
PRACTICE 55:
a) Show that 2(x−4)2+6 is quadratic.
b) Solve 2(x−4)2+6=10.
c) Consider y=2(x−4)2+6. What x-value gives the smallest possible value for y?
- Question What is the best way to solve (2x+3)2=19 ?
Solve (x+1)3=27.
Solve −3x2+5x+2=1 via the box method.
What is the perimeter of a rectangle of area 25 square inches with one side 2 inches longer than then other?
Solving (x−a)(x−b)=0 obviously gives x=a orx=b. But if we expand the equation it reads: x2−(a+b)x+ab=0. Apply the box method to x2−(a+b)x+ab=0. Does it also give x=a or x=b?
Factorising
Let's have a look at a slightly more complex quadratic equation
This equation contains an x-term (target) as well as an x^2 term, which means that our previous method of isolating x^2 on one side and then taking square roots will no longer work.
But there is a different trick to help us - we can factorise one "x" out of both
Now we can use a useful property of multiplication: if the product of two terms is 0, then one of the two terms must also be zero. There is no way you can get 0 by multiplying two numbers which are both not 0.
Image
In our example, this means that either
Exercises
Here is another quadratic equation that can be solved using factoring:
Unlike before, we cannot just factor out x, because we'd still have the 5 at the end left over. Our solution needs to be a bit more clever:
If you expand those brackets, you will find that it is exactly the same. But now we can use the same trick for a product that is 0, to find that the quadratic equation has two solutions:
Unfortunately, this doesn't explain how we found two numbers 2 and 3 that just happened to work in the equation above. To work that out, we can work backwards:
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Let's start by revising how to simplify terms with brackets. For example The trick is simply to add up all possible pairs of numbers, while taking care to respect all minus signs:
Now, if we have a quadratic equation like
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Finding the numbers P and Q always takes a little bit of guesswork, but in all the examples below it should be relatively straightforward.
Exercises!
Try to find the missing number in these factorisation examples:
x^2 + 3x + 2 = (x+1)(x+
Some quadratic equations look completely ordinary to start with, but when we factorise them, we're only left with a single bracket: In these cases there is just a single solution for the quadratic equation.
And finally, some quadratic equations actually have a coefficient in front of . This makes the factorisation a bit more difficult, but it still works the same way: