Question

Find the roots of each of the following quadratic equations if they exist by the method of completing the squares:
√5x2 + 9x + 4√5 = 0

Answer 1

Question:

Find the roots of the following quadratic equations, if they exist by the method of completing the square: √5x² + 9x + 4√5 = 0

Answer:

x = -√5 , -4√5/5

Note:

• The possible values of unknown (variable) for which the equation is satisfied are called its solutions or roots .

• If x = a is a solution of any equation in x , then it must satisfy the given equation otherwise it's not a solution (root) of the equation.

• The discriminant of the the quadratic equation

ax² + bx + c = 0 , is given as ; D = b² - 4ac

• If D > 0 then its roots are real and distinct.

• If D < 0 then its roots are imaginary.

• If D = 0 then its roots are real and equal.

Solution:

Here,

The given quadratic equation is :

√5x² + 9x + 4√5 = 0

Clearly, here we have ;

a = √5

b = 9

c = 4√5

Now,

The discriminant will be ;

=> D = b² - 4ac

=> D = 9² - 4•√5•4√5

=> D = 81 - 80

=> D = 1 ( D > 0 )

Since,

The discriminant of the given quadratic equation is greater than zero, thus there must exist two distinct real roots.

Now,

=> √5x² + 9x + 4√5 = 0

Dividing both sides by √5 , we have ;

=> x² + 9x/√5 + 4 = 0

=> x² + 9x/√5 + (9/2√5)² - (9/2√5)² + 4 = 0

=> x² + 2•x•(9/2√5) + (9/2√5)² = (9/2√5)² - 4

=> (x + 9/2√5)² = 81/20 - 4

=> (x + 9/2√5)² = (81 - 80)/20

=> (x + 9/2√5)² = 1/20

=> x + 9/2√5 = √(1/20)

=> x + 9/2√5 = ± 1/2√5

=> x = - 9/2√5 ± 1/2√5

Case1

=> x = -9/2√5 + 1/2√5

=> x = (-9+1)/2√5

=> x = -8/2√5

=> x = -4/√5

=> x = -4√5/5

Case2

=> x = -9/2√5 - 1/2√5

=> x = (-9-1)/2√5

=> x = -10/2√5

=> x = -5/√5

=> x = -√5

Hence,

The roots of the given quadratic equation are :

x = -√5 , -4√5/5