It is known that the equation #bx^2-(a-3b)x+b=0# has one real root. Prove that the equation #x^2+(a-b)x+(ab-b^2+1)=0# has no real roots.?
2 Answers
See below.
Explanation:
The roots for
The roots will be coincident and real if
or
Now solving
The condition for complex roots is
now making
Concluding, if
We are given that the equation:
# bx^2-(a-3b)x+b=0 #
has one real root, therefore the discriminant of this equation is zero:
# Delta = 0 #
# => (-(a-3b))^2 - 4(b)(b) = 0 #
# :. (a-3b)^2 - 4b^2 = 0 #
# :. a^2-6ab+9b^2 - 4b^2 = 0 #
# :. a^2-6ab+5b^2 = 0 #
# :. (a-5b)(a-b) = 0 #
# :. a=b# , or#a=5b #
We seek to show the equation:
# x^2+(a-b)x+(ab-b^2+1) = 0 #
has no real roots. This would require a negative discriminant. The discriminant for this equation is:
# Delta = (a-b)^2 - 4(1)(ab-b^2+1) #
# \ \ \ = a^2-2ab+b^2 -4ab+4b^2-4 #
# \ \ \ = a^2-6ab+5b^2-4 #
And now let us consider the two possible cases that satisfy the first equation:
Case 1:
# Delta = a^2-6ab+5b^2-4 #
# \ \ \ = (b)^2-6(b)b+5b^2-4 #
# \ \ \ = b^2-6b^2+5b^2-4 #
# \ \ \ = -4 #
# \ \ \ lt 0 #
Case 2:
# Delta = a^2-6ab+5b^2-4 #
# \ \ \ = (5b)^2-6(5b)b+5b^2-4 #
# \ \ \ = 25b^2-30b^2+5b^2-4 #
# \ \ \ = -4 #
# \ \ \ lt 0 #
Hence the conditions of the first equation are such that the second equation always has a negative discriminant, and therefore has complex roots (ie no real roots), QED