1
Question
Solve the following equations :
xa+λ+yb+λ+zc+λ=1,xa+μ+yb+μ+zc+μ=1,xa+ν+yb+ν+zc+ν=1.
xa+λ+yb+λ+zc+λ=1,xa+μ+yb+μ+zc+μ=1,xa+ν+yb+ν+zc+ν=1.
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Solution
Consider the following equation in t,
xa+t+yb+t+zc+t=1−(t−λ)(t−μ)(t−γ)(t+a)(t+b)(tc)
x,y,z being for the present regarded as know quantities.
This equation when cleared of fractions is of second degree in t, and is satisfied for three values t=λ,t=μ and t=v by virtue of the given equations; hence it must be an identity.To find the value of x, we multiply (1) by a+t, and then put t=−a. Thus
x=(a+λ)(a+μ)(a+v)(a−b)(a−c)
or x=(−a−λ)(−a−μ)(−a−v)(−a+b)(−a+c).
By reason of symmetry, we have
y=(b+λ)(b+μ)(b+v)(b−c)(b−a)
and z=(c+λ)(c+μ)(c+v)(c−a)(c−b)
We know that if α,β are the roots of an eqution, then this equation is
x2−(α+β)x+αβ=0
or x2−S1x+S2=0
where S1 is sum of roots taken one at a time and S2 is sum of products of roots taken two at a time.Similarly if there is an equation whose roots are αβγ then S1=α+β+γ,S2=αβ+βγ+γα,S3=αβγ and the corresponding equation will be t3−S1t2+S2t−S3=0.
Again we know that if a is a root of the equation f(x)=0 then a will satisfy the equation, i.e.f(a)=0.
Alternative Solution of Q.22 (i).
Consider the equation z+ty+t2x+t3=0
It is clear that the above equation is satisfied by t=a,b,c by virtue of the above three given equations.which is true by 1st equation and so on. Thus a,b,c are the roots of the equation.
∵z+ay+a2x+a3=0.
which is ture by 1st equation and so on. Thus a,b,c are the roots of the equation.
t3+t2x+ty+z=0
∴S1=a+b+c=−x,
S2=ab+bc+ca=y,S3=abc=−z
∴x=−(a+b+c),y=∑ab,z=−abc.
xa+t+yb+t+zc+t=1−(t−λ)(t−μ)(t−γ)(t+a)(t+b)(tc)
x,y,z being for the present regarded as know quantities.
This equation when cleared of fractions is of second degree in t, and is satisfied for three values t=λ,t=μ and t=v by virtue of the given equations; hence it must be an identity.To find the value of x, we multiply (1) by a+t, and then put t=−a. Thus
x=(a+λ)(a+μ)(a+v)(a−b)(a−c)
or x=(−a−λ)(−a−μ)(−a−v)(−a+b)(−a+c).
By reason of symmetry, we have
y=(b+λ)(b+μ)(b+v)(b−c)(b−a)
and z=(c+λ)(c+μ)(c+v)(c−a)(c−b)
We know that if α,β are the roots of an eqution, then this equation is
x2−(α+β)x+αβ=0
or x2−S1x+S2=0
where S1 is sum of roots taken one at a time and S2 is sum of products of roots taken two at a time.Similarly if there is an equation whose roots are αβγ then S1=α+β+γ,S2=αβ+βγ+γα,S3=αβγ and the corresponding equation will be t3−S1t2+S2t−S3=0.
Again we know that if a is a root of the equation f(x)=0 then a will satisfy the equation, i.e.f(a)=0.
Alternative Solution of Q.22 (i).
Consider the equation z+ty+t2x+t3=0
It is clear that the above equation is satisfied by t=a,b,c by virtue of the above three given equations.which is true by 1st equation and so on. Thus a,b,c are the roots of the equation.
∵z+ay+a2x+a3=0.
which is ture by 1st equation and so on. Thus a,b,c are the roots of the equation.
t3+t2x+ty+z=0
∴S1=a+b+c=−x,
S2=ab+bc+ca=y,S3=abc=−z
∴x=−(a+b+c),y=∑ab,z=−abc.
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