COMPLETING THE SQUARE PRACTICE WORKSHEET

Solve the following quadratic equations by completing the square

(i) x2+6x–7  =  0

(ii)  x2+3x+1  =  0

(iii)  2x2+5x-3  =  0

(iv)  4x2+4bx–(a2-b2)  =  0

(v)  (5x+7)/(x–1)  =  3x+2

Solution

Question 1 :

x2+6x–7  =  0

Solution :

x2+6x–7  =  0

Step 1 :

Notice that the coefficient of x2 is 1, so don't have to divide the entire equation by any numerical value.

Step 2 :

Write the coefficient of x as multiple of 2.

x2+2x3 –7  =  0

Step 3 :

Here the first and second term is in the form a2 + 2ab, in order to complete this formula, we need b2.

Here b  =  3

x2+2x3+32-32 –7  =  0

(x+3)2–9–7  =  0

(x+3)2–16  =  0

(x+3)2  =  16

(x+3)  =  √16

x+3  =  ±4

x+3  =  4

x  =  1

x+3  =  -4

x  =  -7

Question 2 :

x2+3x+1  =  0

Solution :

Step 1 :

Notice that the coefficient of x2 is 1, so don't have to divide the entire equation by any numerical value.

Step 2 :

Write the coefficient of x as multiple of 2.

x2+2x⋅(3/2)+1  =  0

Step 3 :

Here the first and second term is in the form a2 + 2ab, in order to complete this formula, we need b2.

Here b  =  3

 x22x⋅(3/2)+(3/2)2-(3/2)2+1  =  0

[x + (3/2)]2 – (9/4) + 1 = 0

[x+(3/2)]2  =  (9/4) – 1

[x+(3/2)]2  =  5/4

x+(3/2)  =  √(5/4)

[x+(3/2)]  =  ±√5/2

x+(3/2)  =  √5/2

x  =  (√5-3)/2

x+(3/2)  =  -√5/2

x  =  (-√5-3)/2

So, the solution is {(√5-3)/2, (-√5-3)/2}.

Question 3 :

2x2+5x-3  =  0

Solution :

Step 1 :

Notice that the coefficient of x2 is 2, so divide the entire equation by 2.

x2+(5x/2)-(3/2)  =  0

Step 2 :

Write the coefficient of x as multiple of 2.

x2+2x⋅(3/2)+1  =  0

Step 3 :

Here the first and second term is in the form a2 + 2ab, in order to complete this formula, we need b2.

Here b  =  3

2x² + 5x - 3 = 0

Now we are going to divide the whole equation by 2

(x+5/4)2  =  49/16

(x+5/4)  =  49/16

(x+5/4)  =  ±7/4

x+5/4  =  7/4

x  =  2/4

x  =  1/2

x+5/4  =  -7/4

x  =  -12/4

x  =  -3

So, the solution is {-3, 1/2}.

Question 4 :

4x2+4bx–(a2 - b2)  =  0

Solution :

4x2+4bx–(a2 - b2)  =  0

Now we are going to divide the whole equation by 4

x2+bx–[(a2 - b2)/4]  =  0

x2+2(x)(b/2)+(b/2)2–(b/2)2 – [(a2 - b2)/4]  =  0

[x + (b/2)]2 – (b2/4) – [(a2-b2)/4]  =  0

[x + (b/2)]2  =  (b2/4)+[(a2-b2)/4]

[x + (b/2)]2  =  [(b2+a2-b2)/4]

[x + (b/2)]2  =  a2/4

[x + (b/2)]  =  √(a²/4)

[x + (b/2)]  =  ±(a/2)

x + (b/2)  =  a/2

x  =  (a-b)/2

x + (b/2)  =  -a/2

x  =  (-a-b)/2

So, the solution is {(a-b)/2, (-a-b)/2}.

Question 5 :

(5x+7)/(x–1)  =  3x+2

Solution :

(5x+7)  =  (3x+2)(x–1)

5x+7  =  3x2–x–2

3x2–x–5x–7-2  =  0

3x2–6x–9  =  0

x2–2x–3  =  0

(x–3)(x+1)  =  0

x-3  =  0

x  =  3

x+1  =  0

x  =  -1

So, the solution is {-1, 3}.

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