Find the zeros of the polynomial x^2 + 8x + 16

Mathematics

1 answer:

Scorpion4ik [409]2 years ago

4 0

hey mate,

$x^2 + 8x + 16 = 0$

<=>

$(x+4)^2 = 0$

As we know that $(a+b)^2 = a^2+b^2+2ab$

then

$(x+4)^2 = 0$

<=> (x+4) = 0

<=> x = -4

so there is one solution x = -4

thanks

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Vector upper a overscript right-arrow endscripts has a magnitude of 69 units and points due west, while vector upper b overscrip

Tom [10]

Refer to the figure below.

Define unit vectors as follows:
$\hat{i}$ in the eastern direction
$\hat{j}$ in the northern direction

Given:
Vectors a and b point due west and south respectively. Therefore
$\vec{a} = -69 \ht{i} \\
\vec{b} = -69\hat{j}$

Part (a)
$\vec{a}+\vec{b}=-69\hat{i}-69\hat{j}=-69(\hat{i}+\hat{j})$
The magnitude is
|-69(√[1² + 1²])| = 69√2 = 97.58.
The direction relative to west is
θ = tan⁻¹ (b/a) = tan⁻¹ 1 = 45° south of west

Answer:
The magnitude of (a + b) is 97.58 or 69√2.
The direction is 45° south of west.

Part (b).
$\vec{a} - \vec{b} = -69\hat{i}+69\hat{j}=-69(\hat{i} -\hat{j})$
The magnitude is
|69(√(1+1)| = 69√2 = 97.58.
The direction relative to west is
tan⁻¹ (b/a) = 45° north of west.

Answer:
The magnitude of (a-b) is 97.58 or 69√2.
The direction is 45° north of west.