Find all the zeros
x^4+3x^2-54=0

Ziros is : x=+-sqrt(6) and x=+-3i

Step-by-step explanation:

x^4+3x^2-54=0

t=x^2, then

t^2+3t-54=0

a=1, b=3, c=-54 (Quadratic equation coefficients)

t1 = (-b+sqrt(b^2-4ac) /2a

t1 = (-3+sqrt(3^2 - 4*1*(-54))/2*1

t1 = (-3+sqrt(9+216))/2

t1 = (-3+sqrt(225))/2

t1 = (-3+15)/2

t1 = 12/2

t1 =6

t2 = (-b-sqrt(b^2-4ac)/2a

t2 =(-3-15)/2

t2=(-18)/2

t2=-9

t1=x^2 =6

x=+-sqrt(6)

t2=x^2=-9

x=+-sqrt(-9)... (-1=i^2)

x=+-sqrt(9i^2)

x=+-3i

the value of v.w= -12.

step-by-step explanation:

we are given three vector r,v and w as:

r = < 5, -5, -2> ; v = < 2, -8, -8> ; w = < -2, 6, -5>

we know that r,v and w could also be represented as:

r=5i-5j-2k

v=2i-8j-8k

and w= -2i+6j-5k

where i,j and k are the unit vectors of x,y and z-axis respectively.

also we know that it has the following property:

i.i=1 ; j.j=1; k.k=1

and i.=j.i=i.k=k.i=j.k=k.j=0

i.e. when we multiply two vectors or we find it's dot product we need to only multiply the components of x coordinates with each other; y with each other and z with each other.

hence, v.w= (2i-8j- -2i+6j-5k)

                  = 2×(-2)+(-8)×(6)+(-8)×(-5)

                  = -4-48+40

                  = -12

hence, v.w= -12

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