suppose r and s are the two soplutions to the eqution x^2+3x-12=0 what is r^2+s^2?

Question

please show your work. i will choose the best explanation as the best answer

Answer 1

Using the quadratic equation we get:
r = -3/2 + sqrt(9 + 4*12) / 2
r = -3/2 + sqrt(57) / 2
r^2 = 9/4 + 57/4 + sqrt(57)

The other root is the same, just with a minus sign.
s = -3/2 - sqrt(57) / 2
s^2 = 9/4 -sqrt(57) + 57/4

When we add the squares, the sqrt(57) cancel out.
r^2 + s^2 = 9/2 + 57/2
r^2 + s^2 = 66/2
r^2 + s^2 = 33

What's more interesting is to do this using the general terms a, b, c.

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

When we square them:
r^2 = b^2/4a^2 + sqrt(b^2 - 4ac)/a + (b^2 - 4ac)/4a^2
s^2 = b^2/4a^2 - sqrt(b^2 - 4ac)/a + (b^2 - 4ac)/4a^2

Now add and again the sqrt goes away:
r^2 + s^2 = 2(b^2/4a^2) + 2(b^2 - 4ac)/4a^2

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

Now you have a general equation:
r^2 + s^2 = (b^2 - 2ac) / a^2

Double-checking against our prior result...
a = 1
b = 3
c = -12
= 9 - 2*1*(-12) / 1
= 9 + 24
= 33

Answer 2

The easiest way to do this is as follows

Let it roots be α and β
Let a, b, c being the coefficients of ax² + bx + x = 0
So a = 1, b = 3, c = -12
Let it roots be r and s

Then r + s = -b/a = -3/1 = -3
rs = c/a = -12/1 = -12

Now r² + s² = (r + s)² - 2rs
= (-3)² - 2 * -12
= 9 + 24
= 33

This comes from the fact that for polynomials

f(x) = a0 x^ + a1x^(n - 1) + a2x^(n - 2) + .... + an

with zeros α, β, γ, δ, ε, ....

Σα = -a1/a0, Σαβ = a2/a0, Σαβγ = -a3/a0, ----

Answer 3

You would use the quadratic formula to figure out what r and s equal.

x=(-b±√(b^2-4ac))/(2a)

x=(-3±√(3^2-4*1*-12))/(2*1)

x=(-3±√(9+48))/2

x=(-3±√*57))/2

x=(-3+7.54983443527075)/2 and x=(3-7.54983443527075)/2

x= 2.27491721763537 and x= -2.27491721763537

These are the factors of x or r and s.

r=2.27491721763537 and s=-2.27491721763537

(2.27491721763537)^2+ (-2.27491721763537)^2

= 10.3504966941878

Answer 4

r= (-b-root(b^2-4ac))/2a
s= (-b+root(b^2-4ac)/2a
hence, r+s = -(b/a)
and r*s= (b^2-b^2+4ac)/4a^2= c/a
thus, r^2 + s^2 = (r+s)^2-2rs = (b/a)^2 -2(c/a)
here a=1 , b=3 , c= -12, taking coefficients of the eqn.
hence ans. is (3/1)^2 -2(-12/1) = 9+24 = 33

Answer 5

solve for x using the quadratic formula and you'll get two different answers. These are r and s. Then form r^2+s^2.