Question

Find the roots of the quadratic equation if they exist by Completely the square x^2 + 3x - 9 = 0

Answer 1

Given, quadratic equation is x² + 3x - 9 = 0

x² + 3x - 9 = 0

=> x² + 3x = 9

=> x² + 2.(3/2).x = 9

=> x² + 2.(3/2).x + (3/2)² = (3/2)² + 9

=> (x + 3/2)² = 9/4 + 9 = 45/4

taking square root both sides,

=> (x + 3/2) = ±3√5/2

=> x = -3/2 ± 3√5/2 = (-3 ± 3√5)/2

hence, roots are (-3 - 3√5)/2 and (-3 + 3√5)/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 + 3x - 9 = 0

=> x² + 3x = 9  

=> x² + 3x + (3/2)² = (3/2)² + 9

=> x² + 2×(3/2)× x + (3/2)²= 9 + (3/2)²

[a² +2ab + b² = (a+b)²]

=> (x + 3/2)² = 9/4 + 9  

=> (x + 3/2)² = (9+36)/4  

=> (x + 3/2)² = 45/4  

=> (x + 3/2) = √(45/4)

=> (x + 3/2) = √(9×5/4)

=> (x + 3/2) = ±3√5/2  

=> x = -3/2 ± 3√5/2

=>x = (-3 ± 3√5)/2  

Hence, roots of the quadratic equation are (-3 - 3√5)/2 and (-3 + 3√5)/2.

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