Math, asked by TbiaSupreme, 1 year ago

Find the roots of the quadratic equation if they exist by Completely the square x² + 7x - 13 = 0

Answers

Answered by abhi178
1
given quadratic equation is x² + 7x - 13 = 0

x² + 7x - 13 = 0

=> x² + 7x = 13

=> x² + 2.(7/2)x = 13

=> x² + 2.(7/2).x + (7/2)² = 13 + (7/2)²

=> (x + 7/2)² = 13 + 49/4

=> (x + 7/2)² = (52 +49)/4 = 101/4

=> (x + 7/2) = ±√(101)/2

=> x = -7/2 ± √(101)/2

=> x = (-7 ± √101)/2

hence, roots of quadratic equation are (-7 - √101)/2 and (-7 + √101)/2
Answered by nikitasingh79
1

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 + 7x - 13 = 0

=> x² + 7x = 13  

=> x² + 7x +(7/2)² = 13 +(7/2)²

=> x² + 2×(7/2)x +(7/2)² =  13 +(7/2)²  

=> (x + 7/2)² = 13 + 49/4  

=> (x + 7/2)² = (52 +49)/4 = 101/4  

=> (x + 7/2)² = 101/4

=> (x + 7/2) = ±√(101)/2  

=> x = -7/2 ± √(101)/2  

=> x = -7/2 + √101/2

=> x = -7/2 - √101 /2  

Hence, roots of quadratic equation are (-7 - √101)/2 & (-7 + √101)/2  

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