Question

Find the roots of the following quadratic equations, if they exist by the method of completing the square:
5x2 – 6x – 2 = 0

Answer 1

Answer:

x²-6/5x-2/5=0

x²-6/5x+(3/5)²=2/5+(3/5)²

(x-3/5)²=19/25

x-3/5=√19/5

x=√19/5+3/5

Answer 2

Question:

Find the roots of the following quadratic equations, if they exist by the method of completing the square: 5x² - 6x - 2 = 0

Answer:

x = (3+√19)/5 , (3-√19)/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 - 6x - 2 = 0

Clearly, here we have ;

a = 5

b = -6

c = -2

Now,

The discriminant will be ;

=> D = b² - 4ac

=> D = (-6)² - 4•5•(-2)

=> D = 36 + 20

=> D = 56. ( D > 9 )

Since,

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

Now,

=> 5x² - 6x - 2 = 0

Dividing both sides of by 5 , we have ;

=> x² - 6x/5 - 2/5 = 0

=> x² - 6x/5 + (3/5)² - (3/5)² - 2/5 = 0

=> x² - 6x/5 + (3/5)² = (3/5)² + 2/5

=> x² - 2•x•(3/5) + (3/5)² = 9/25 + 2/5

=> (x - 3/5)² = (9+10)/25

=> (x - 3/5)² = 19/25

=> x - 3/5 = √(19/25)

=> x - 3/5 = ± √19/5

=> x = 3/5 ± √19/5

=> x = (3±√19)/5

=> x = (3+√19)/5 , (3-√19)/5

Hence,

The roots of the given quadratic equation are:

x = (3+√19)/5 , (3-√19)/5