467 Quadratic Equations This shows ...
Question

# 467 Quadratic Equations This shows that both the roots of the given equation are real. For equal roots, we must have: D = 0. Now, D=0 = (a - b)2 + (b -c)2+(c-a)2 = 0 = (a - b) = 0, (b-c) = 0 and (c-a) = 0 > a = b = c. Hence, the roots are equal only when a = b = c. AMPLE 12 If the roots of the equation (b-c)x2 + (c-a)x + (a - b) = 0 are equal, prove that 2b = a +c. [CBSE 2002C, '06] UTION Clearly, x = 1 satisfies the given equation. Since its roots are equal, so 1 and 1 are its roots. .. product of roots of the given equation = (1 x 1) = 1. But, product of roots = %=6 [ product of roots = Àl : 0= = 1= a-b=b-c = 26 = a+c. Hence, 2b = a +c. a -b h - C LE 13 Show that the equation 3x2 + 7x + 8 = 0 is not true for any real value of x. N The given equation is 3x2 + 7x + 8 = 0. .: D = (72 – 4 x 3 x 8) = (49 - 96) = -47 < 0. So, the given equation has no real roote

11th - 12th Class
Maths
Solution
208
4.0 (1 ratings)
12 ( 2=0 quad f_{0} R ) equal Resty ( (-9)^{2}-4(6-6)(9-6)=0 ) ( left(9^{2}+e^{2}-29 cright)-4left(96-9 c-6^{2}+6 cright)=0 ) ( 9^{2}+46^{2}+c^{2}-496-46 c+29 c=0 ) ( (9-26+c)^{2}=0 ) So ( a-2 b+c=0 ) ( a+c=2 b )
Quick and Stepwise Solutions Just click and Send OVER 20 LAKH QUESTIONS ANSWERED Download App for Free