Question

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.?

Answer 1

See below.

Explanation:

The roots for `bx^2-(a-3b)x+b=0` are

`x = (a - 3 b pmsqrt[a^2 - 6 a b + 5 b^2])/(2 b)`

The roots will be coincident and real if

`a^2 - 6 a b + 5 b^2 = (a - 5 b) (a - b)=0`

or

`a=b` or `a = 5b`

Now solving

`x^2+(a-b)x+(ab-b^2+1)=0` we have

`x = 1/2 (-a + b pm sqrt[a^2 - 6 a b + 5 b^2-4])`

The condition for complex roots is

`a^2 - 6 a b + 5 b^2-4 lt 0`

now making `a = b` or `a = 5b` we have

`a^2 - 6 a b + 5 b^2-4 = -4 < 0`

Concluding, if `bx^2-(a-3b)x+b=0` has coincident real roots then `x^2+(a-b)x+(ab-b^2+1)=0` will have complex roots.

Answer 2

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: `a=b`

`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: `a=5b`

`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