Find the value of k if the equation 4x^2+kx–3=0 has roots –21 and 23?

Question

simplify: [(21x^-2)(y^3) / (15x^3)(y^-5)]

answer: (7x^8) / (5x^2)

The product of two numbers is 24. If one of the numbers is x, express the sum of the two numbers in terms of x.

Find the value of k if the equation 4x^2+kx–3=0 has roots –21 and 23

Please help me with these questions with some steps would be helpful, I am not allowed to use a graphing calculator.

Answer 1

I suppose roots are -21 & 23.
Therefore, (x+21)(x-23)=0
x^2+21x-23x-483=0
x^2-2x-483=0
4x^2-8x-1932=0
comparing it with given equation does not give you any thing. There is somethng wong with the question.

[(21x^-2)(y^3) / (15x^3)(y^-5)]
=(21/15)x^(-2-3) * y^(3+5)
=(7/5)x^-5y^8 or 7y^8/5x^5

(7x^8) / (5x^2)
=7x^(8-2)/5
=(7/5)x^6

other number =24/x

Answer 2

Wait if 4x^2 + kx - 3 = 0
then the product of the roots should be -3/4, but
-21 * 23 = -483.

2. [(21x^-2)(y^3) / (15x^3)(y^-5)] =
= 21x^-2 * y^3 * (x^-3)/15 * y^5 =
21(x^-2)(x^-3)/15 * y^3 * y^5 = 21(x^-5)/15 * y^8 =
= 7/5x^5 * y^8 = (7y^8)/5x^5

3. The product of two numbers is 24. If one of the numbers is x, express the sum of the two numbers in terms of x.

The first number is x, and the second one is 24/x, thus their sum is (x + 24/x)

4. Find the value of k if the equation 4x^2+kx–3=0 has roots –21 and 23

-- Already answered: no solutions.

Answer 3

sum of the roots=-21+23=2
from the eqn. sum of roots will be -k/4
so -k/4=2
k=-8

[(21x^-2)(y^3) / (15x^3)(y^-5)
=7y^8/5x^5

on number is x then the other will be 24/x
sum of the two will be x+24/x
=(x^2+24)/x

Answer 4

aaaaaaa

Answer 5

k=8

all you have to do is divide 21 and -23 with 4x^2+kx–3 using synthetic division.. (synthetic division is very important.. you'd have to know how to do it first.. it's really quite easy..)

since they are roots, they should both result to an equal value, which is zero..

to find k, just solve for it in the equation(from synthetic division):
21k+1761=2113-23k
k=8


and there you have it... ^_^