# 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