QUADRATIC EQUATION PRACTICE PROBLEMS

Solve the equation.

Problem 1 :

3(x + 7)² - 17 = 25

Solution :

3(x + 7)² - 17 = 25

3(x + 7)² = 25 + 17

3(x + 7)² = 42

(x + 7)² = 42/3

(x + 7)² = 14

Take square root on both sides,

x + 7 = ±√14

x = - 7 ± √14

Problem 2 :

3(x + 7)² + 17 = 59

Solution :

3(x + 7)² + 17 = 59

3(x + 7)² = 59 - 17

3(x + 7)² = 42

(x + 7)² = 42/3

(x + 7)² = 14

Take square root on both sides,

x + 7 = ±√14

x = - 7 ± √14

Problem 3 :

Solve the equation 4x² + 20 = 0.

Solution:

4x² + 20 = 0

4x² = 20

x² = 5

x = ±√5

x = √5 (or) -√5

Problem 4 :

Solve the equation 4x² + 5 = -7

Solution :

4x² + 5 = -7

4x² = -7 - 5

4x² = -12

x² = -3

x = √-3

x = i√3

Problem 5 :

Find the value of c that makes x² - 4x + c a perfect square trinomial.

Write the new expression as the square of a binomial.

Solution :

It satisfies the condition b² - 4ac = 0 and are in the form of ax² + bx + c.

As per question, 

a = 1, b = -4, and c = c

b² - 4ac = 0

(-4)² - 4(1)c = 0

16 - 4c = 0

- 4c = -16

c = 4

So, the required equation is x² - 4x + 4.

By square of binomial,

= (x - 2)²

Problem 6 :

Jenny is solving the equation x² - 8x = 9 by completing the square. What number should be added to both sides of the equation to complete the square?

a)  2    b)  4   c)  8   d)  16

Solution :

x² - 8x = 9

Add to both sides the square of half of the coefficient of x that is 16.

x² - 8x + 16 = 9 + 16

(x - 4)² = 25

Take square root of both sides,

x - 4 = 5

x = 5 + 4

x = 9

Hence, 16 should be added to both sides of the equation to complete the square.

Problem 7 :

What are the solutions to the equation x² + 2x + 2 = 0?

Solution:

x² + 2x + 2 = 0

Since the quadratic equation is not factored, we are using the quadratic formula.

x = -b ± √(b² - 4ac)/2a

a = 1, b = 2, and c = 2

x = -2 ± √[2² - 4(1)(2)]/2

x = [-2 ± √-4] / 2

x = -2 ± 2i / 2

x = -1 ± i

x = -1 + i and -1 - i

Problem 8 :

What are the solutions to the equation 1 + 1/x² = 3/x?

Solution :

1 + 1/x² = 3/x

Find common denominator.

(x² + 1) / x² = 3/x

Multiplying x² on both sides, 

x² ∙ (x² + 1) / x² = (3/x) ∙ x²

x² + 1 = 3x

x² - 3x + 1 = 0

Use the quadratic formula,

x = -b ± √ (b² - 4ac) / 2a

a = 1, b = -3, and c = 1

x = 3 ± √ [(-3)² - 4(1)(1)] / 2

x = 3 ± √5 / 2

Problem 9 :

There are two numbers with the following properties.

1. The second number is 3 more than the first number.
2. The product of the two numbers is 9 more than their sum.

Which of the following represents possible values of these two numbers?

a)  -6, -3   b)  -4, -1   c)  -1, 4   d)  -3, 6

Solution :

y = x + 3 ---> (1)

xy = x + y + 9 ---> (2)

Substitute y = x + 3 in (2)

x(x + 3) = x + (x + 3) + 9

x² + 3x = 2x + 12

x² + 3x - 2x - 12 = 0

x² + x - 12 = 0

By factorization,

(x - 3) (x + 4) = 0

x = 3, -4

If x = -4, then 

y = -4 + 3

y = -1

So, option (B) is correct.