Menu Close

a-b-c-are-the-roots-from-equation-x-3-5x-2-9x-10-0-If-P-x-Ax-3-Bx-2-Cx-2015-and-P-a-b-c-P-b-a-c-P-c-a-b-What-is-the-value-of-A-B-C-

Tinku Tara 0 Comments
Pin




Question Number 11753 by Joel576 last updated on 31/Mar/17
a, b, c are the roots from equation x^3 − 5x^2 − 9x + 10 = 0 If P(x) = Ax^3 + Bx^2 + Cx − 2015 and P(a) = b + c P(b) = a + c P(c) = a + b What is the value of A + B + C ?
a,b,caretherootsfromequationx3−5x2−9x+10=0IfP(x)=Ax3+Bx2+Cx−2015andP(a)=b+cP(b)=a+cP(c)=a+bWhatisthevalueofA+B+C?
Answered by sma3l2996 last updated on 31/Mar/17
(x−a)(x−b)(x−c)=x^3 −5x^2 −9x+10=0 (x^2 −(a+b)x+ab)(x−c)=x^3 −(a+b+c)x^2 +(ab+ac+bc)x−abc a+b+c=5 , ab+ac+bc=−9 , abc=−10 we have P(a)+P(b)+P(c)=2(a+b+c)=2×5 P(a)+P(b)+P(c)=A(a^3 +b^3 +c^3 )+B(a^2 +b^2 +c^2 )+C(a+b+c)−3×2015 we have: =a^3 +b^3 +c^3 +3(a^2 b+a^2 c+b^2 c+bc^2 +ab^2 +ac^2 )+6abc so: a^3 +b^3 +c^3 =(a+b+c)^3 −3(a(ab+ac)+b(bc+ab)+c(bc+ac))−6abc =5^3 −3(a(−9−bc)+b(−9−ac)+c(−9−ab))−6×(−10) =125+3(9a+abc+9b+abc+9c+abc)+60 =3(9(a+b+c)+3abc)+185 a^3 +b^3 +c^3 =27×5+9×(−10)+185=230 we have also (a+b+c)^2 =a^2 +b^2 +c^2 +2(ab+ac+bc) so: a^2 +b^2 +c^2 =(a+b+c)^2 −2(ab+ac+bc)=5^2 −2(−9) a^2 +b^2 +c^2 =43 so: P(a)+P(b)+P(c)=230A+43B+5C−6045=2×5=10 230A+43B+5C=6055
(x−a)(x−b)(x−c)=x3−5x2−9x+10=0(x2−(a+b)x+ab)(x−c)=x3−(a+b+c)x2+(ab+ac+bc)x−abca+b+c=5,ab+ac+bc=−9,abc=−10wehaveP(a)+P(b)+P(c)=2(a+b+c)=2×5P(a)+P(b)+P(c)=A(a3+b3+c3)+B(a2+b2+c2)+C(a+b+c)−3×2015wehave:=a3+b3+c3+3(a2b+a2c+b2c+bc2+ab2+ac2)+6abcso:a3+b3+c3=(a+b+c)3−3(a(ab+ac)+b(bc+ab)+c(bc+ac))−6abc=53−3(a(−9−bc)+b(−9−ac)+c(−9−ab))−6×(−10)=125+3(9a+abc+9b+abc+9c+abc)+60=3(9(a+b+c)+3abc)+185a3+b3+c3=27×5+9×(−10)+185=230wehavealso(a+b+c)2=a2+b2+c2+2(ab+ac+bc)so:a2+b2+c2=(a+b+c)2−2(ab+ac+bc)=52−2(−9)a2+b2+c2=43so:P(a)+P(b)+P(c)=230A+43B+5C−6045=2×5=10230A+43B+5C=6055
Commented by sma3l2996 last updated on 31/Mar/17
working...
working…
Commented by Joel576 last updated on 01/Apr/17
thank you very much
thankyouverymuch
Answered by ajfour last updated on 31/Mar/17
A+B+C = 2625
A+B+C=2625
Commented by ajfour last updated on 31/Mar/17
As a,b, c are roots of x^3 −5x^2 −9x+10=0 or (x+2)(x^2 −7x+5) =0 are x=−2 , ((7±(√(29)))/2) we take a=−2; b+c=7; bc=((49−29)/4)= 5 ; a+b+c=5 . P(a)=b+c=Aa^3 +Ba^2 +Ca−2015 7=−8A+4B−2C−2015 .....(i) P(b)=a+c=Ab^3 +Bb^2 +Cb−2015 P(c)=a+b=Ac^3 +Bc^2 +Cc−2015 P(b)−P(c) gives b−c= (b−c)[A(b^2 +bc+c^2 ) +B(b+c)+C ] so, A[(b+c)^2 −bc] +B(b+c)+C=−1 ......(1) 44A+7B+C=−1 .....(ii) similarly [P(a)−P(b)]/(a−b) yields A(a^2 +ab+b^2 )+B(a+b)+C=−1 ......(2) and [P(c)−P(a)]/(c−a) gives A(c^2 +ac+a^2 )+B(c+a)+C=−1 ......(3) Adding eqns (1), (2), and (3) A{ 2(a^2 +b^2 +c^2 )+(ab+bc+ca)} +2B(a+b+c)+3C=−3 A{2(a+b+c)^2 −3(ab+bc+ca)} +2B(a+b+c)+3C=−3 or A{50−3(−9)}+10B+3C=−3 or 77A+10B+3C=−3 ...(iii) solving (i), (ii), and (iii) I found A+B+C = 2625 .
Asa,b,carerootsofx3−5x2−9x+10=0or(x+2)(x2−7x+5)=0arex=−2,7±292wetakea=−2;b+c=7;bc=49−294=5;a+b+c=5.P(a)=b+c=Aa3+Ba2+Ca−20157=−8A+4B−2C−2015…..(i)P(b)=a+c=Ab3+Bb2+Cb−2015P(c)=a+b=Ac3+Bc2+Cc−2015P(b)−P(c)givesb−c=(b−c)[A(b2+bc+c2)+B(b+c)+C]so,A[(b+c)2−bc]+B(b+c)+C=−1……(1)44A+7B+C=−1…..(ii)similarly[P(a)−P(b)]/(a−b)yieldsA(a2+ab+b2)+B(a+b)+C=−1……(2)and[P(c)−P(a)]/(c−a)givesA(c2+ac+a2)+B(c+a)+C=−1……(3)Addingeqns(1),(2),and(3)A{2(a2+b2+c2)+(ab+bc+ca)}+2B(a+b+c)+3C=−3A{2(a+b+c)2−3(ab+bc+ca)}+2B(a+b+c)+3C=−3orA{50−3(−9)}+10B+3C=−3or77A+10B+3C=−3…(iii)solving(i),(ii),and(iii)IfoundA+B+C=2625.
Commented by Joel576 last updated on 01/Apr/17
thank you very much
thankyouverymuch

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *