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Class 10 > If the zeroes of polynomial x3 -- ax2 + bx -- c are...
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If the zeroes of polynomial x3 – ax2 + bx – c are in AP then show that 2a3 – 9ab + 27c = 0

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zreors of polynomial means roots of polynomial if the roots of polynomial are in AP then the roots are A+d, A, A-d Sum of the roots = (-) Co-efficient of X^2/ Co-efficient of X^3 = A+d +A +A-d -(-a) = 3A a =3A A= a/3 since...
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zreors of polynomial means roots of polynomial if the roots of polynomial are in AP then the roots are A+d, A, A-d Sum of the roots = (-) Co-efficient of X^2/ Co-efficient of X^3 = A+d +A +A-d -(-a) = 3A a =3A A= a/3 since f(x)= x^3 - ax^2 +bx - c f(A)= A^3 - aA^2 +bA -c put A= a/3 (a/3)^3 -a(a/3)^2 +b(a/3) -c=0 (a^3)/27 - (a^3)/9 +ab/3 -c = 0 (a^3) -3(a^3) +9ab -27c =0 -2(a^3) +9ab - 27c = 0 2a^3 -9ab +27c = 0 hence PROVED read less
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let l , m , n are zeroes of given polynomial. now sum of l+m+n = a product lmn = c product of zeros taken two at time lm+mn+nl = b also it is given that zeros of polynomial are in A. P . then 2m= l+ n from these equations we get a = 3m , b=2m2 + ln. now putting values...
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let l , m , n are zeroes of given polynomial. now sum of l+m+n = a product lmn = c product of zeros taken two at time lm+mn+nl = b also it is given that zeros of polynomial are in A. P . then 2m= l+ n from these equations we get a = 3m , b=2m2 + ln. now putting values of a, b , c in LHL we get 0. read less
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Let p-1, p, p+1 be the zeroes of the given polynomial. p-1+p+p+1=a 3p=a (p-1)p+p(p+1)+(p-1)(p+1)=b p2-p+p2+p+p2-1=b 3p2-1=b (p-1)p(p+1)=c (p2-1)p=c p3-p=c Consider 2a3-9ab+27c 2(3p)3-9(3p)(3p2-1)+27(p3-p)= 54p3-81p3+27p+27p3-27p= 81p3-81p3=0 Therefore 2a3-9ab+27c=0
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Let the zeros of given polynomial is m-d, m, and m+d. Here m and d are not zero. substituting the value of zeros in the polynomial, we have=>(m-d)^3-a(m-d)^2+b(m-d)-c=0 =>m^3-d^3-3m^2b+3md^2-a*(m^2+d^2-2*m*d)+b*m-b*d-c=0 =>m^3-d^3-3m^2b+3md^2-a*m^2-ad^2+2*a*m*d+b*m-b*d-c=0----------------(i) Similarily,...
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Let the zeros of given polynomial is m-d, m, and m+d. Here m and d are not zero. substituting the value of zeros in the polynomial, we have=>(m-d)^3-a(m-d)^2+b(m-d)-c=0 =>m^3-d^3-3m^2b+3md^2-a*(m^2+d^2-2*m*d)+b*m-b*d-c=0 =>m^3-d^3-3m^2b+3md^2-a*m^2-ad^2+2*a*m*d+b*m-b*d-c=0----------------(i) Similarily, we get a new equation by putting the (m+d), =>m^3+d^3+3m*d^2-a*m^2-a*d*d^2-2a*m*d+b*m+b*d-c=0---------------------(ii) Again, Putting the value of m in polynomial, we have m^3-a*m^2+b*m-c=0-----------------------------------------(iii) Adding equation (i) & (ii), we have=>2*m^3+6*m*d^2-2a*m^2-2a*d^2+2b*m-2*c=0 =>2*(3^3-a*m^2+b*m-c)+6m*d^2-2*a*d^2=0 =>6*m*d^2=2*a*d^2 =>m=a/3------------------------------(iv) Now, Putting the value of m in the polynomial, =>(a/3)^3-a*(a/3)^2+b*a/3-c=0 =>a^3/27-a^3/9+a*b/3-c=0 =>a^3-3*a^3+9*a*b-27*c=0 =>2*a^3-9*a*b+27*c=0 (Proved) read less
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As we know, sum of roots = -(coeff. of x2)/(coeff. of x3) = a Since the roots ( or zeros) of polynomial are in AP, then we can assume the roots to be A-D, A, A+D ( with D as common diff.) sum of roots = A-D+A+A+D = 3A =a therefore A = a/3 Since, A is a zero of the polynomial, then it should satisfy...
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As we know, sum of roots = -(coeff. of x2)/(coeff. of x3) = a Since the roots ( or zeros) of polynomial are in AP, then we can assume the roots to be A-D, A, A+D ( with D as common diff.) sum of roots = A-D+A+A+D = 3A =a therefore A = a/3 Since, A is a zero of the polynomial, then it should satisfy the eqn. put A=a/3 in eqn (a^3)/27 - (a^3)/9 + (a*b)/3 - c = 0 Hence, 2a3-9ab+27c = 0 read less
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