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Find the non-zero value of $a$ such that the quadratic equation $ax^2+8x+4=6x-40$ has only one solution.

 Sep 11, 2023

Best Answer 

 #2
avatar+36923 
+1

CPhill made just a slight error :

Re-write as  ax^2 + 2x + 46  = 0      Should be    ax^2 + 2x + 44   

 

Then following through with his calculations will show a = 4/ 176 =   1/44 

 Sep 11, 2023
edited by ElectricPavlov  Sep 11, 2023
 #1
avatar+129376 
+1

Re-write as  ax^2 + 2x + 46  = 0 

 

If we only have one solution , the discriminant =  0

 

So

 

2^2 - 4(a) (46)  = 0

 

4 - 184a =  0

 

4 = 184a

 

a = 4/184  = 1 / 46

 

cool cool cool

 Sep 11, 2023
 #2
avatar+36923 
+1
Best Answer

CPhill made just a slight error :

Re-write as  ax^2 + 2x + 46  = 0      Should be    ax^2 + 2x + 44   

 

Then following through with his calculations will show a = 4/ 176 =   1/44 

ElectricPavlov Sep 11, 2023
edited by ElectricPavlov  Sep 11, 2023

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