x^2=2x (3k+1)-7 (2k+3)

Question

Find the value (s) of k which make the Quadratic have 2 identical Real roots 

Answer 1

2 identical real roots ==> discriminant = b^2 - 4ac = 0  

 

Put the quadratic in standard form  ax^2 + bx + c 

 

x^2 - 2(3k+1)x + 7(2k+3)  

 

In this form  a = 1 , b = -2(3k+1) , and  c = 7(2k+3)  so  b^2 - 4ac  is  

 

[-2(3k+1)]^2 - 4*(1)*[7(2k+3)] = 0

4*(9k^2 + 6k + 1) - 4*[14k + 21] = 0

9k^2 + 6k + 1 - 14k - 21 = 0

9k^2 - 8k - 20 = 0 

(9k + 10)*(k - 2) = 0  

9k + 10 = 0  OR  k - 2 = 0

k = -10/9  OR k = 2  

 

You can check by plugging these values in for k and factoring the resulting quadratic  

 

k = 2:  x^2 - 14x + 49 = (x-7)*(x-7) has 2 identical real roots at x = 7  

k = -10/9:  x^2 +(14/3)x + (49/9) = [x+(7/3)]*[x+(7/3)] has 2 identical real roots at x = -7/3