Hello,

P(x)=x^4-6x²+2=(x-a)(x-b)(x-c)(x-d)
=x^4-(a+b+c+d)x^3+(ab+ac+ad+bc+bd+cd)x^2-(abc+abd+acd+bcd)x+abcd

==>
ab+ac+ad+bc+bd+cd=-6
abc+abd+acd+bcd=0
abcd=2
a+b+c+d=0 ==>(a+b+c+d)²=0=a²+b²+c²+d²+2(ab+ac+ac+bc+bd+cd)
==>a²+b²+c²+d²=0-2*(-6)=12

if a is a root P(a)=0==>a^4-6a²+2=0
if b is a root P(b)=0==>b^4-6b²+2=0
if c is a root P(a)=0==>c^4-6c²+2=0
if d is a root P(a)=0==>d^4-6d²+2=0

==>a^4+b^4+c^4+d^4-6(a²+b²+c²+d²)+4*2=0
==>a^4+b^4+c^4+d^4=-8+6*12=64

6 0

3 years ago

A booster club sells raffle tickets • Before tickets go on sale to the public, 120 tickets are sold to student athletes. • After

GaryK [48]

Answer:

The top graph.

Step by step:

Before the tickets go on sale, 120 tickets were already sold.

After that, the tickets sell at a constant rate for a total of 8 hours.

At the end of this time, all the tickets were sold (we have 0 tickets left)

If x (horizontal axis) represents the hours since tickets go on sale, and y (vertical axis) represents the number of raffle tickets sold.

Then, at x = 0, we should already see y = 120

Because we start with 120 tickets sold.

Then, as x increases, the number of tickets sold also should increase, until we get x = 8 hours, where y stops increasing because all tickets are already sold.

Then we should have an increasing line that stops increasing at x = 8 hours.

Then the correct option is the above graph, where we have:

An increasing line.

y = 120 in the vertical axis (y = 120 when x = 0)

3 0

3 years ago