How do you solve x^4-16x^2-720=0

Factorise a) 3x^2-7x+2 b)2x^2-x-3 c)3x^2-16x-12

Luba_88 [7]

Answer: Answers are in the steps read carefully!

Step-by-step explanation:

A) 3x^2 - 7x + 2 To factor this polynomial, you have to find two numbers that their product is 6 and their sum is -7. The numbers -1 and -6 works out because -6 times -1 is 6 and -6 plus -1 is -7.

Now rewrite the polynomial as

3x^2 - 1x - 6x + 2 Now group it

(3x^2 - 1x) (-6x+2) Factor it by groups

x (3x -1) -2(3x -1) Now factor out 3x-1

(3x -1) (x-2) Done!

B) 2x^2 - x -3 Now the same way.You will have two numbers that their product is -6 and their sum is -1. You may be wondering how I get -6 .I get -6 by multiply the leading coefficient 2 by the constant -3. The numbers -3 and 2 works out. Because -3 times 2 is -6 and -3 plus 2 is -1.

Rewrite the polynomial as

2x^2 +2x - 3x -3 GRoup them and factor them

(2x^2 + 2x) (-3x-3)

2x(x+1) -3(x+1) Factor out x+1

(x+1) (2x -3) Done!

C) 3x^2 - 16x - 12 Find two numbers that their product is -36 and their sum is -12. The numbers -18 and 2 works out because -18 times 2 is -36 and -18 plus 2 is -16.

Rewrite the polynomial

3x^2 +2x -18x - 12 GRoup them

(3x^2 + 2x) (-18x - 12) Factor them

x (3x +2) -6(3x +2) Factor out 3x+2

(3x+2) (x -6) Done !

6 0

3 years ago

Which expression gives the magnitude of the magnetic field in the region c < r2 < b (at e?

BaLLatris [955]

The electric currents and magnetic materials have an effect called magnetic field. This magnetic field is specified with its direction and magnitude. The magnitude is the strength of the field.

the magnetic field at a distance r from a long straight wire carrying current I can be calculated with the equation:

B = μ I / 2 pi R

where μ is a constant, I is the current and R is the radius.

5 0

2 years ago

Looking at the top of tower A and base of tower B from points C and D, we find that ∠ACD = 60°, ∠ADC = 75° and ∠ADB = 30°. Let t

katrin2010 [14]

Answer:

$\text{Exact: }AB=25\sqrt{6},\\\text{Rounded: }AB\approx 61.24$

Step-by-step explanation:

We can use the Law of Sines to find segment AD, which happens to be a leg of $\triangle ACD$ and the hypotenuse of $\triangle ADB$ .

The Law of Sines states that the ratio of any angle of a triangle and its opposite side is maintained through the triangle:

$\frac{a}{\sin \alpha}=\frac{b}{\sin \beta}=\frac{c}{\sin \gamma}$

Since we're given the length of CD, we want to find the measure of the angle opposite to CD, which is $\angle CAD$ . The sum of the interior angles in a triangle is equal to 180 degrees. Thus, we have:

$\angle CAD+\angle ACD+\angle CDA=180^{\circ},\\\angle CAD+60^{\circ}+75^{\circ}=180^{\circ},\\\angle CAD=180^{\circ}-75^{\circ}-60^{\circ},\\\angle CAD=45^{\circ}$

Now use this value in the Law of Sines to find AD:

$\frac{AD}{\sin 60^{\circ}}=\frac{100}{\sin 45^{\circ}},\\\\AD=\sin 60^{\circ}\cdot \frac{100}{\sin 45^{\circ}}$

Recall that $\sin 45^{\circ}=\frac{\sqrt{2}}{2}$ and $\sin 60^{\circ}=\frac{\sqrt{3}}{2}$ :

$AD=\frac{\frac{\sqrt{3}}{2}\cdot 100}{\frac{\sqrt{2}}{2}},\\\\AD=\frac{50\sqrt{3}}{\frac{\sqrt{2}}{2}},\\\\AD=50\sqrt{3}\cdot \frac{2}{\sqrt{2}},\\\\AD=\frac{100\sqrt{3}}{\sqrt{2}}\cdot\frac{ \sqrt{2}}{\sqrt{2}}=\frac{100\sqrt{6}}{2}={50\sqrt{6}}$

Now that we have the length of AD, we can find the length of AB. The right triangle $\triangle ADB$ is a 30-60-90 triangle. In all 30-60-90 triangles, the side lengths are in the ratio $x:x\sqrt{3}:2x$ , where $x$ is the side opposite to the 30 degree angle and $2x$ is the length of the hypotenuse.

Since AD is the hypotenuse, it must represent $2x$ in this ratio and since AB is the side opposite to the 30 degree angle, it must represent $x$ in this ratio (Derive from basic trig for a right triangle and $\sin 30^{\circ}=\frac{1}{2}$ ).