+0  
 
0
30
3
avatar

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

 Sep 6, 2023

Best Answer 

 #2
avatar+177 
+1

The given quadratic equation is \(ax^2 + 8x + 4 = 6x - 40\). First, simplify it to standard form.

 

\(ax^2 + 8x + 4 = 6x - 40 \\ ax^2 + 2x + 44 = 0\)

 

Now, we can use the discriminant of a quadratic, which gives information about the number of solutions. In order for the quadratic to have only one solution, the discriminant should be equal to 0.

 

\(D = 0 \\ b^2 - 4ac = 0 \\ 2^2 - 4 * a * 44 \\ 4 - 176a = 0 \\ 176a = 4 \\ a = \frac{4}{176} \\ a = \frac{1}{44}\)

 Sep 7, 2023
 #1
avatar+12 
0

 

 

(2x + 2)(2x + 2) = 0   so   a = 4 

 Sep 6, 2023
edited by SoloX  Sep 6, 2023
 #2
avatar+177 
+1
Best Answer

The given quadratic equation is \(ax^2 + 8x + 4 = 6x - 40\). First, simplify it to standard form.

 

\(ax^2 + 8x + 4 = 6x - 40 \\ ax^2 + 2x + 44 = 0\)

 

Now, we can use the discriminant of a quadratic, which gives information about the number of solutions. In order for the quadratic to have only one solution, the discriminant should be equal to 0.

 

\(D = 0 \\ b^2 - 4ac = 0 \\ 2^2 - 4 * a * 44 \\ 4 - 176a = 0 \\ 176a = 4 \\ a = \frac{4}{176} \\ a = \frac{1}{44}\)

The3Mathketeers Sep 7, 2023
 #3
avatar+36923 
+1

Nice work, The3Mathketeers ! 

ElectricPavlov  Sep 7, 2023

4 Online Users

avatar
avatar
avatar
avatar