Hey there! We receieved your request
Stay Tuned as we are going to contact you within 1 Hour
One of our academic counsellors will contact you within 1 working day.
Click to Chat
1800-5470-145
+91 7353221155
Use Coupon: CART20 and get 20% off on all online Study Material
Complete Your Registration (Step 2 of 2 )
Sit and relax as our customer representative will contact you within 1 business day
OTP to be sent to Change
Given equation is
(x - b)(x - c) + (x - c)(x - a) + (x - a)(x - b) = 0
On simplification, equation changes to
x2 - (b + c)x + bc + x2 - (c + a)x + ca + x2 - (a +b)x + ab = 0
3x2 - (2a + 2b + 2c)x + (ab + bc + ca) = 0
3x2 - 2(a + b + c)x + (ab + bc + ca) = 0
On comparing the above equation with Ax2 + Bx + C = 0,
A = 3, B = -2(a + b + c), C = (ab + bc + ca)
Since roots are equal
Discriminant D = 0
B2 - 4AC = 0
[-2(a + b + c)]2 - 4(3)(ab + bc + ca) = 0
4(a + b+ c)2 -12(ab + bc + ca) = 0
4(a2 + b2 + c2 + 2ab + 2bc + 2ca) - 12ab -12bc -12ca = 0
4a2 + 4b2 + 4c2 + 8ab + 8bc + 8ca - 12ab -12bc -12ca = 0
4a2 + 4b2 + 4c2 - 4ab -4bc -4ca = 0
4(a2 + b2 + c2 - ab -bc -ca) = 0
Dividing both sides by 4
a2 + b2 + c2 - ab -bc -ca= 0
Hence answer is option D - None of these
(x-b)(x-c)+(x-c)(x-a)+(x-a)(x-b)=0
On simplification, we get
3x^2 -2(a+b+c)x + (ab+bc+ca) = 0.
Roots are equal implying that D = 0.
i.e 4(a+b+c)^2 -4(3)(ab+bc+ca) = 0
or, 4[a^2 + b^2 +c^2 +2(ab+bc+ca) -3(ab+bc+ca)]=0
or, 4[a^2 + b^2 + c^2 -ab-bc-ca]=0
or, 2[2a^2 + 2b^2 +2c^2 - 2ab -2bc-2ca]=0
or, (a^2 + b^2 -2ab) + (b^2 + c^2 -2bc) + (c^2 + a^2 -2ac)=0
i.e (a-b)^2 + (b-c)^2 + (c-a)^2 = 0
This is true only if all three terms are zero.
Hence a=b=c.
Now option B becomes a( 1+omega + omega(squared) ) = a*0 = 0.
Hence option B is correct.
Get your questions answered by the expert for free
You will get reply from our expert in sometime.
We will notify you when Our expert answers your question. To View your Question
Win Gift vouchers upto Rs 500/-
Register Yourself for a FREE Demo Class by Top IITians & Medical Experts Today !