+0  
 
0
56
2
avatar

Let a be a real number for which there exists a unique value of b such that the quadratic equation x^2 + 2bx + (a-b) = 0 has one real solution. Find a.

 

Thank you everyone that Helps!, but please Clearly say your Answer!

 Mar 25, 2020

Best Answer 

 #2
avatar+20899 
+1

I plan to use the quadratic formula on this equation:   x2 + 2bx + (a - b)  =  0  

using     a = 1     b = 2b     c = a - b  

For the equation to have only one real solution  b2 - 4ac  must be zero.

Replacing:     (2b)2 - 4(1)(a - b)  =  0

                             4b2 - 4a + 4b  =  0

                                             -4a  =  -4b2 - 4b

Diviing by -4:                           a  =  b2 + b

 

I hope that the variables in the equation did not confuse you with the variables in the quadratic formula.

 Mar 25, 2020
 #1
avatar+1956 
+1

Hint:

 

One root means that the equation has to be in the form of (x+#)^2=0.

 

Hope this helped!

 Mar 25, 2020
 #2
avatar+20899 
+1
Best Answer

I plan to use the quadratic formula on this equation:   x2 + 2bx + (a - b)  =  0  

using     a = 1     b = 2b     c = a - b  

For the equation to have only one real solution  b2 - 4ac  must be zero.

Replacing:     (2b)2 - 4(1)(a - b)  =  0

                             4b2 - 4a + 4b  =  0

                                             -4a  =  -4b2 - 4b

Diviing by -4:                           a  =  b2 + b

 

I hope that the variables in the equation did not confuse you with the variables in the quadratic formula.

geno3141 Mar 25, 2020

8 Online Users

avatar