Worked examples
The question
Question: choose a method and solve x² = 49.
x2 = 49
1 Choose the method
There is no x term.
So we can get x² alone and take the square root.
That is the quickest way.
So: square root.
1. no x term, or ( )² = k? ✓ square root
2 Square root of both sides
x squared is 49.
So x is the square root of 49, or minus it.
Minus times minus is plus, so the negative one works too.
x² = k → x = √k or x = −√k
7 × 7 = 49, (−7) × (−7) = 49
7 times 7 is 49.
So x = 7 or x = −7.
x = 7 or x = −7
Careful!
Many people write only x = 7.
They lose the minus answer.
Careful: not only x = 7 ✗
3 Check
Let's check: put x = 7 into the left side.
x = 7: 72 = 49
We get 49, the same as the right side.
It works!
Now put x = −7 into the left side.
x = −7: (−7)2 = 49
We get 49 again, the same as the right side.
So both answers are right!
Let's look back.
First: choose the method.
Then: square root of both sides.
Last: check.
▶ Watch it solved
The question
Question: choose a method and solve 3x² − 27 = 0.
3x2 − 27 = 0
1 Choose the method
There is no x term.
So we can get x² alone and take the square root.
That is the quickest way.
So: square root.
1. no x term, or ( )² = k? ✓ square root
2 Get x² alone
There is no x term, so we get x² by itself.
To remove the − 27, we add 27 on both sides.
3x2 − 27 + 27 = 0 + 27
0 + 27 = 27.
Now we have 3x² = 27.
3x2 = 27
To remove times 3, we divide both sides by 3.
3x23 = 273
27 divided by 3 is 9.
x2 = 9
3 Square root of both sides
x squared is 9.
So x is the square root of 9, or minus it.
Minus times minus is plus, so the negative one works too.
x² = k → x = √k or x = −√k
3 × 3 = 9, (−3) × (−3) = 9
3 times 3 is 9.
So x = 3 or x = −3.
x = 3 or x = −3
Careful!
Many people write only x = 3.
They lose the minus answer.
Careful: not only x = 3 ✗
4 Check
Let's check: put x = 3 into the left side.
x = 3: 3(3)2 − 27 = 27 − 27 = 0
We get 0, the same as the right side.
It works!
Now put x = −3 into the left side.
x = −3: 3(−3)2 − 27 = 27 − 27 = 0
We get 0 again, the same as the right side.
So both answers are right!
Let's look back.
First: choose the method.
Then: get x² alone.
Then: square root of both sides.
Last: check.
▶ Watch it solved
The question
Question: choose a method and solve x² = 4x + 21.
x2 = 4x + 21
1 Choose the method
We ask in order.
First: is there no x term?
No, there is one.
1. no x term, or ( )² = k? ✗
x² has no number in front, and two whole numbers work.
So it factors.
So: factoring.
2. factors easily? ✓ factoring
2 Get 0 on one side
To solve it, we want 0 on the right side.
So we take away 4x + 21 on both sides.
We do it on both sides, so they stay equal.
x2 − 4x − 21 = 4x + 21 − 4x − 21
Now we have x² − 4x − 21 = 0.
x2 − 4x − 21 = 0
3 Factor
We need two numbers.
They multiply to −21, and they add to −4.
? × ? = −21
? + ? = −4
They multiply to a minus number, so one number is plus and one is minus.
They add to a minus number, so the bigger one is minus.
× is − → one +, one −; bigger one −
We list the pairs that make 21, smallest first.
We take the smaller from the bigger.
We need 4.
3 and 7 give 4.
Yes!
With the signs: 3 and −7.
1 × 21: 21 − 1 = 20
3 × 7: 7 − 3 = 4 ✓
So it is (x + 3)(x − 7) = 0.
(x + 3)(x − 7) = 0
4 One bracket is 0
Two numbers multiply to 0.
That only happens if one of them is 0.
So one bracket must be 0.
A × B = 0 → A = 0 or B = 0
We make each bracket 0.
Two small equations.
x + 3 = 0 or x − 7 = 0
In x + 3 = 0, we take away 3 on both sides.
In x − 7 = 0, we add 7 on both sides.
x + 3 − 3 = 0 − 3 or x − 7 + 7 = 0 + 7
So x = −3 or x = 7.
x = −3 or x = 7
5 Check
Let's check: put x = −3 into the left side.
x = −3, left: (−3)2 = 9
And the right side.
right: 4(−3) + 21 = −12 + 21 = 9
Both sides are 9.
It works!
Now put x = 7 into the left side.
x = 7, left: 72 = 49
And the right side.
right: 4(7) + 21 = 28 + 21 = 49
Both sides are 49.
So both answers are right!
Let's look back.
First: choose the method.
Then: get 0 on one side.
Then: factor.
Then: one bracket is 0.
Last: check.
▶ Watch it solved
