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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 6, 2023

#2
+177
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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
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(2x + 2)(2x + 2) = 0   so   a = 4

Sep 6, 2023
edited by SoloX  Sep 6, 2023
#2
+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.

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

The3Mathketeers Sep 7, 2023
#3
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Nice work, The3Mathketeers !

ElectricPavlov  Sep 7, 2023