Question

Evaluate each of the following expressions for x = -2 , y = -1 , z = 3 :
(i) (x/y) + (y/z) + (z/x)
(ii) xx + yy + zz - xy - yz - zx
​

Answer 1

Form a quadratic equation whose Roots are x,y,z

Form a quadratic equation whose Roots are x,y,z Now, let the equation be ax3+bx2+cx+d=0

Form a quadratic equation whose Roots are x,y,z Now, let the equation be ax3+bx2+cx+d=0 So, Sum x+y+z=−b/a

Form a quadratic equation whose Roots are x,y,z Now, let the equation be ax3+bx2+cx+d=0 So, Sum x+y+z=−b/a xy+yz+xz=c/a

Form a quadratic equation whose Roots are x,y,z Now, let the equation be ax3+bx2+cx+d=0 So, Sum x+y+z=−b/a xy+yz+xz=c/a xyz=−d/a

If a=1 , equation will be x3−6x2+15x−14=0

If a=1 , equation will be x3−6x2+15x−14=0 So, x=2 is a root of equation : the equation can be written as (x−2)(x2−4x+7)=0

If a=1 , equation will be x3−6x2+15x−14=0 So, x=2 is a root of equation : the equation can be written as (x−2)(x2−4x+7)=0 Since x2−4x+7=0 has imaginary roots which are 2+isqrt3 and 2−isqrt3

If a=1 , equation will be x3−6x2+15x−14=0 So, x=2 is a root of equation : the equation can be written as (x−2)(x2−4x+7)=0 Since x2−4x+7=0 has imaginary roots which are 2+isqrt3 and 2−isqrt3 So, the only real solution for this equation is x=2

If a=1 , equation will be x3−6x2+15x−14=0 So, x=2 is a root of equation : the equation can be written as (x−2)(x2−4x+7)=0 Since x2−4x+7=0 has imaginary roots which are 2+isqrt3 and 2−isqrt3 So, the only real solution for this equation is x=2 Hope this was helpful.

If a=1 , equation will be x3−6x2+15x−14=0 So, x=2 is a root of equation : the equation can be written as (x−2)(x2−4x+7)=0 Since x2−4x+7=0 has imaginary roots which are 2+isqrt3 and 2−isqrt3 So, the only real solution for this equation is x=2 Hope this was helpful.Regards.

Answer 2

1) (-2/-1)+(-1/3)=-2-1/3
= -6-1/3
=>-7/3

2)(-2*-2)+(-1*-1)+(3*3)-(-2*-1)-(-1*3)-(3*-2)
=4+1+9-2+3+6
=>21