What is the discriminant of $x^{2} + 8 x + 16 = 0$ $x^2 + 8x + 16 = 0$ and what does that mean?

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What is the discriminant of $x^{2} + 8 x + 16 = 0$ $x^2 + 8x + 16 = 0$ and what does that mean?

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Answer 1 · 2015-07-23T00:16:49

The expression is of the form $A x^{2} + B x + C = 0$ $Ax^2+Bx+C=0$
where $A = 1, B = 6, C = 16$ $A=1,B=6,C=16$

Explanation:

The discriminant is defined as $D = B^{2} - 4 A C$ $D=B^2-4AC$
If $D > 0$ $D>0$ there are two solutions to the equation
If $D = 0$ $D=0$ there is one solution
If $D < 0$ $D<0$ there is no solution (in real numbers)

In your case $D = 8^{2} - 4 \cdot 1 \cdot 16 = 0 \to$ $D=8^2-4*1*16=0->$ one solution.

The equation can be written as ${(x + 4)}^{2} \to x = - 4$ $(x+4)^2->x=-4$

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What is the discriminant of $x^{2} + 8 x + 16 = 0$ $x^2 + 8x + 16 = 0$ and what does that mean?

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