HOW TO SOLVE QUADRATIC EQUATIONS BY COMPLETING THE SQUARE

If possible, solve for x using ‘completing the square’ :

Example 1 :

x² + 2x  =  5

Solution :

The coefficient of x² is 1, so we dont have to divide the entire equation.

x² + 2x  =  5

Half of the coefficient of x(2/2) is 1.

x² + 2x + 1²  =  5 + 1²

x² + 2x + 1  =  6

(x + 1)²  =  6

Taking square roots on both sides, we get

(x + 1)  =  ±√6

So, solution are √6-1 and -√6-1.

Example 2 :

x² - 2x  =  - 7

Solution :

x² - 2x  =  - 7

Coefficient of x² is 1. So dividing the entire equation is not required.

Coefficient of x is -2, half of it  =  -1

x² - 2x + (-1)²  =  - 7 + (-1)²

x² - 2x + 1  =  - 7 + 1

x² - 2x + 1  =  - 6

(x - 1)²  =  - 6

So, the system has no real solution.

Example 3 :

x² - 6x  =  - 3

Solution :

x² - 6x  =  - 3

Coefficient of x² is 1. So dividing the entire equation is not required.

Coefficient of x is -6, half of it  =  -3

x² - 6x + (-3)²  =  - 3 + (-3)²

x² - 6x + 9  =  - 3 + 9

x² - 6x + 9  =  6

(x - 3)²  =  6

(x - 3)  =  ±√6

So, the solutions are √6+3 and -√6+3.

Example 4 :

Solve by completing the square method 

2x² + 5x - 3  =  0

Solution :

2x² + 5x - 3  =  0

2x² + 5x  =  3/2

Divide the equation by 2.

x² + (5/2)x  =  (3/2)

Half of coefficient of x is 5/4.

By adding (5/4)² on both sides, we get 

x² + (5/2)x + (5/4)²  =  (3/2) + (5/4)²

x² + (5/2)x + (5/4)²  =  (3/2) + (5/4)² 

[ x + (5/4) ]²  =  (3/2) + (25/16)

[ x + (5/4) ]²  =  49/16

[ x + (5/4) ]²  =  √(49/16)

x + (5/4) = ± 7/4

So, the solutions area 1/2 and -3.

Example 5 :

The height y (in feet) of a baseball t seconds after it is hit can be modeled by the function

y = −16t² + 96t + 3

Find the maximum height of the baseball. 

Solution :

y = −16t² + 96t + 3

By writing the quadratic function in vertex form, we get after how long the ball will reach the maximum height.

y = −16[t² - 6t] + 3

y = −16[t² - 2t(3) + 3² - 3²] + 3

y = −16[(t - 3)² - 3²] + 3

y = −16[(t - 3)² - 9] + 3

y = −16(t - 8)² + 144 + 3

y = −16(t - 8)² + 147

So, the ball will reach its maximum height at 8 seconds and the maximum height is 147 feet.

Example 6 :

While marching, a drum major tosses a baton into the air and catches it. The height h (in feet) of the baton t seconds after it is thrown can be modeled by the function

h = −16t ² + 32t + 6. 

a. Find the maximum height of the baton.

b. The drum major catches the baton when it is 4 feet above the ground. How long is the baton in the air?

Solution :

h = −16t² + 32t + 6. 

h = −16[t² - 2t] + 6

h = −16[t² - 2t(1) + 1² - 1²] + 6

= −16[(t - 1)² - 1²] + 6

= −16(t - 1)² + 16 + 6

= −16(t - 1)² + 22

So, the maximum height is 22 feet.

b) When the height is 4

4 = −16t² + 32t + 6. 

−16t² + 32t + 6 - 4 = 0

−16t² + 32t + 2 = 0

-16[t² - 2t] + 2 = 0

-16[t² - 2t(1) + 1² - 1²] + 2 = 0

-16[(t - 1)² - 1²] + 2 = 0

-16(t - 1)² + 16 + 2 = 0

-16(t - 1)² = -18

-16(t - 1)² = -18

(t - 1)² = 9/8

(t - 1)² =  ±1.125

t - 1 = 1.125 and t - 1 = -1.125

t = 2.125 and t = 0.125

So, 2 second and 0.125 seconds are answers.

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