17. Completing the square (a >1)

Question 1

LEVEL 8

Complete the square for the following quadratic:

3x^2 - 18x + 10.

Select the correct answer from the options below:

A: 3(x - 3)^2 - 17 = 0

B: 3(x - 3)^2 - 27 = 0

C: 3(x - 6)^2 - 17 = 0

D: 3(x - 6)^2 - 27 = 0

CORRECT ANSWER: A: 3(x - 3)^2 - 17 = 0

WORKED SOLUTION:

Question 2

LEVEL 8

write 3x^2 + 12x + 2 in the form a(x+b)^2 +c, where a, b and c are integers.

Select the correct answer from the options below:

A: 3(x + 2)^2 - 2

B: (3x + 2)^2 +10

C: 3(x + 2)^2 + 10

D: 3(x + 2)^2 - 10

CORRECT ANSWER:    D: 3(x + 2)^2 - 10

WORKED SOLUTION:

Factorise first: 3[x^2 + 4x] + 2

Complete the square: 3[(x + 2)^2 - 4] + 2

Rearrange: 3(x + 2)^2 - 12 + 2

Question 3

LEVEL 8

write 2x^2 – 12x + 7 in the form a(x+b)^2 +c, where a, b and c are integers.

Select the correct answer from the options below:

A: 2(x - 3)^2 - 12

B: 2(x - 3)^2 - 11

C: 2(x - 3)^2 - 9

D: 2(x - 3)^2 + 9

CORRECT ANSWER:   B: 2(x - 3)^2 - 11

WORKED SOLUTION:

Question 4

LEVEL 8

Complete the square for the following quadratic:

3q^2 + 15q - 10

Select the correct answer from the options below:

A: 3\left(q + \frac{5}{2}\right)^2 - \frac{115}{12}

B: 3\left(q + \frac{5}{2}\right)^2 + \frac{115}{4}

C: 3\left(q + \frac{5}{2}\right)^2 - \frac{115}{4}

D: 3\left(q + \frac{15}{2}\right)^2 - \frac{115}{4}

CORRECT ANSWER:   C: 3\left(q + \frac{5}{2}\right)^2 - \frac{115}{4}

WORKED SOLUTION:

Having a coefficient of the squared term being bigger than 1 can makes completing the square tricky.

When fractions get involved it makes it even more difficult.

Factorise first: 3[q^2 + 5q] - 10

Complete the square: 3[(q + 2.5)^2 - 6.25] - 10

Rearrange: 3(q + 2.5)^2 - 18.75 - 10

Which is

Question 5

LEVEL 8

write 2x^2 + 4x - 8 in the form a(x+b)^2 +c, where a, b and c are integers.

Select the correct answer from the options below:

A: 2(x + 2)^2 - 5

B: 2(x + 2)^2 - 5

C:  2(x + 1)^2 - 10

D: 2(x + 1)^2 - 5

CORRECT ANSWER:     C:  2(x + 1)^2 - 10

WORKED SOLUTION:

Factorise first: 2[x^2 + 2x] - 8

Complete the square: 2[(x + 1)^2 - 1] - 8

Rearrange: 2(x + 1)^2 - 2 - 8

Question 6

LEVEL 8

write 5x^2 + 10x + 15 in the form a(x+b)^2 +c, where a, b and c are integers.

Select the correct answer from the options below:

A:  5 (x + 1)^2 + 10

B: 5 (x + 1)^2 + 5

C: 5 (x + 5)^2 + 10

D: 5 (x + 5)^2 + 5

CORRECT ANSWER:    A:  5 (x + 1)^2 + 10

WORKED SOLUTION:

Factorise first: 5[x^2 + 2x] + 15

Complete the square: 5[(x + 1)^2 - 1] + 15

Rearrange: 5(x + 1)^2 - 5 + 15

Question 7

LEVEL 8

Write 2y^2-16y+6 in the form a(x+b)^2+c, where a, b, and c are constants to be determined.

Select the correct answer from the options below:

A:  2(y-4)^2 - 26

B: 2(y-4)^2 - 13

C: (2y-4)^2 - 26

D: 2(y-8)^2 - 122

CORRECT ANSWER:    A:  5 (x + 1)^2 + 10

WORKED SOLUTION:

We need to start by taking a factor of 2 out of the whole expression to get it in the form we are use to. Doing so, we get

2y^2-16y+6=2(y^2-8y+3)

Now, all we need to do is complete the square on the bit inside the bracket (as we normally would) and then bring the 2 back in at the end. So, we get

y^2-8y+3 = (y-4)^2+3-(-4)^2=(y-4)^2-13

Remember, this is just the inside of the bracket, so if we bring the 2 back into the picture, we get

2\left[(y-4)^2-13\right]

Finally, expanding the square brackets (and only those), we get the final form of the expression to be

2(y-4)^2 - 26

This is precisely the form the question asked for.

Question 8

LEVEL 8

Use completing the square to solve x^2 + 4x + 3 = 0.

Select the correct answer from the options below:

A: x = -1 and x = 3

B: x = 1 and x = -3

C: x = -1 and x = -3

D: x = 1 and x = 3

CORRECT ANSWER:   C: x = -1 and x = -3

WORKED SOLUTION:

Using the completing the square method our equation becomes

(x + 2)^2 + 3 - 2^2 = (x + 2)^2 - 1 = 0

Now, to solve this equation for x we will rearrange our equation – not expanding the brackets – to make x the subject. Firstly, add 1 to both sides to get

(x + 2)^2 = 1

Then, square root both sides to get

x + 2 = \pm\sqrt{1} = \pm 1

Finally, subtracting 2 from both sides makes x the subject and leaves us with

x = \pm 1 - 2.

Therefore, the two solutions to our quadratic are x = -1 and x = -3.

Note: Make sure you remember to include both the positive and negative square roots, or you will end up missing one of the solutions.