if the roots of the quadratic equation :
(x-a) (x-b) + (x-b) (x-c) + (X-c) (x-a) =0 are equal, then show that a = b = c
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Answers
Answer:
a = b = c
Step-by-step explanation:
Given Equation is (x - a)(x - b) + (x - b)(x - c) + (x - c)(x - a) = 0
⇒ [x² - (a + b)x + ab] + [x² - (b + c)x + bc] + [x² - (c + a)x + ac] = 0
⇒ [3x² - 2(a + b + c)x + ab + bc + ca] = 0
On comparing with ax² + bx + c = 0. we get
a = 3, b = -2(a + b + c), c = ab + bc + ca
∴ D = b² - 4ac
= [-2(a + b + c)² - 4(3)(ab + bc + ca]
= 4(a + b + c)² - 12(ab + bc + ca)
= 4[(a + b + c)² - 3(ab + bc + ca)]
= 4[a² + b² + c² + 2ab + 2bc + 2ca - 3ab - 3bc - 3ca]
= 4[a² + b² + c² - ab - bc - ca]
= 2[2a² + 2b² + 2c² - 2ab - 2bc - 2ca]
= 2[(a - b)² + (b - c)² + (c - a)²]
Here,D ≥ 0, if D = 0, then
⇒ [(a - b)² + (b - c)² + (c - a)²] = 0
⇒ a - b = 0, b - c = 0, c - a = 0
⇒ a = b = c.
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Step-by-step explanation:
,(x-a)(x-b)+(x-b)(x-c)+(x-c)(x-a)=0
x^2-(a+b)x+ab+x^2-(b+c)x+bc+x^2-(c+a)x+ca=0
3x^2–2(a+b+c)x+(ab+bc+ca)=0
Roots are equal , B^2–4AC=0
[-2(a+b+c)]^2 -4×3×(ab+bc+ca)=0
4(a+b+c)^2=4×3×(ab+bc+ca)
(a+b+c)^2=3(ab+bc+ca)
a^2+b^2+c^2+2(ab+bc+ca)=3(ab+bc+ca)
a^2+b^2+c2=ab+bc+ca……………..(1)
Now put a=b=c = (k) let
k^2+k^2+k^2=kk+kk+kk
3k^2=3k^2 , (true) . Proved.