If x+y+z=1 and x^2+y^2+z^2=3, then find the value of xy+xz+yz?

Answer 1

(x+y+z)^2 = (x+y+z)(x+y+z)= x^2+y^2+z^2+2xy+2xz+2yz =>
1*1=3+2(xy+xz+yz)=> 2(xy+xz+yz)= 1-3 =>
xy+xz+yz =-2/2 => xy+xz+yz = -1

Answer 2

Man, all these variables must give you a headache!

So how could we make things a bit easier.
How about Pick & Choose Numbers.
Note that the problem doesn't ask for ALL x, y and z so you can find an x, y, and z for which the given conditions are true and go from there.

Given:
x + y + z = 1
x^2 + y^2 + z^2 = 3
what is xy + xz + yz?

so let's pick
x = +1
y = -1
z = +1

so
x + y + z = 1
(1) + (-1) + (1) = 1?
check!

x^2 + y^2 + z^2 = 3
(1)^2 + (-1)^2 + (1)^2 = 3
1 + 1 + 1 = 3?
check!

so
xy + xz + yz is
(1)(-1) + (1)(1) + (-1)(1)
(-1) + (1) + (-1) ---> -1

xy + xz + yz = -1

Answer 3

Here is the working out:
(a) x + y + z = 1 (given)
(b) x**2 + y**2 + z**2 = 3 (given)

then
(c) x + y = 1 - z (reorder a)
(d) x**2 + y**2 + 2xy = 1 - 2z + z**2 (squaring both sides of c)
(e) z**2 - 2xy = 2 + 2z - z**2 (b minus d)
(f) 2z**2 - 2xy = 2 + 2z (reorder e)
(g) z**2 - xy = 1 + z (divide f by 2)
similarly we can derive h and i:
(h) y**2 - xz = 1 + y
(i) x**2 - yz = 1 + x
then
(j) x**2 + y**2 + z**2 - xy - xz - yz = 3 + x + y + z (g plus h plus i)
(k) 3 - xy - xz - yz = 3 + 1 (substituting a and b into j)
(l) xy + xz + yz = 1 (reorder k)

So the answer is: xy+xz+yz = 1

Answer 4

By expanding (x+y+z)^2 and substituting x^2+y^2+z^2=3,
we get xy+xz+yz = -1

Answer 5

i visit in person-friendly words teach that there aren't any integer strategies. can could imagine on it some extra. x^2 + y^2 + z^2 = 6 + yz + xz + xy Rearrange to quadratic kind in x : x^2 - x(y + z) + (y^2 + z^2 -yz - 6) = 0 opt for for x utilizing the quadratic formula : x = {y + z ± sqrt[(-(y + z))^2 - 4(y^2 + z^2 - yz - 6)]} / 2 This simplifies to : x = {y + z ± sqrt[3(8 - (y - z)^2)]} / 2 Now the expression lower than the sq. root could be better than or equivalent to 0. i.e. 8 - (y - z)^2 ? 0 i.e 8 ? (y - z)^2 i.e. (y - z)^2 ? 8 the in person-friendly words circumstances that fulfill this are y - z = 0 or ± a million or ± 2. Substituting those into sqrt[3(8 - (y - z)^2)] supplies sqrt(24) or sqrt(21) or sqrt(12), none of which yield an integer. to that end, there aren't any integer strategies.

Answer 6

Given,x+y+z=1 & x^2+y^2+z^2=3

(x+y+z)^2=x^2+y^2+z^2+2xy+2yz+2xz
1^2=3+2xy+2yz+2xz
3-2-2-2=xy+yz+xz
1-4=xy+yz+xz
-3=xy+yz+xz
ie,xy+yz+xz=(-3).

Answer 7

x + y +z = 1
x^2 + y^2 + z^2 =3

(x+y+z)^2 = 1^2
x^2 + xy + xz + xy +y^2 + yz + xz + yz + z^2 = 1
x^2 + y^2 + z^2 + 2xy + 2yz + 2xz = 1
3 + 2(xy+yz+xz) = 1
2(xy+yz+xz) = -2
xy + yz + xz = -2/2
so, xy + yz + xz = -1

Answer 8

3

Answer 9

x+y+z=1 so (x+y+z)^2=1^2=1

Thus
(x+y+z)^2= x^2+y^2+z^2+2(xy+yz+xz) =3+2(xy+xz+yz)=1^2=1

Thus xy+yz+xz=-2/2=-1