Question

Find the roots of the quadratic equation if they exist by Completely the square x² –x – 3 = 0

Answer 1

Given quadratic equation is x² – x – 3 = 0

x² - x - 3 = 0

=> x² - x = 3

=> x² + (-1)x = 3

=> x² + 2.(-1/2).x = 3

=> x² + 2.(-1/2).x + (-1/2)² = (-1/2)² + 3

if a = x , b = -1/2 then, x² + 2.(-1/2).x + (-1/2)² = a² + 2ab + b² = (a + b)² [ from algebraic identities]
so, x² + 2.(-1/2).x + (-1/2)² = {x + (-1/2)}²

=> {x + (-1/2)}² = 1/4 + 3

=> (x - 1/2)² = (13)/4

taking square root both sides,

=> x - 1/2 = ±√13/2

=> x = 1/2 ± √13/2 = (1 ±√13)/2

hence, roots are (1 + √13)/2 and (1 - √13)/2

Answer 2

METHOD OF COMPLETING THE SQUARE :  

Step 1 - Write the given equation in standard form, ax²+bx+c = 0, a≠0.

Step 2 -  If the coefficient of x² is 1, go to step 3. If not, divide both sides of the equation by the coefficient of  x²

Step 3 -  Shift the constant term (c/a) on RHS.

Step 4- Find half the coefficient of x and square it. Add this number to both sides of  the equation.  

Step 5 - Write LHS in the form a perfect square and simplify the RHS.

Step 6 - Take the square root on both sides.

step 7 : Find the values of x by shifting the constant term(b/2a) on RHS from LHS.

SOLUTION :  

Given :

x² - x - 3 = 0

=> x² - x = 3  

=> x² (-1)x + (½)² = 3 +(½)²

=> x² - 2× (1/2) × x  + (½)² = 3 + (½)²

[a² - 2ab + b² = (a - b)² ]

=>(x  -½ )² = 1/4 + 3

=>(x  -½ )² = (1+ 12)/4  

=> (x - ½ )² = 13/4  

=> x - 1/2 = ±√13/2  

=> x = 1/2 ± √13/2

x = (1 ±√13)/2  

Hence, roots of the quadratic equation are (1 + √13)/2 and (1 - √13)/2

HOPE THIS ANSWER WILL HELP YOU...