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.?

Oct 17, 2017

See below.

Explanation:

The roots for $b {x}^{2} - \left(a - 3 b\right) x + b = 0$ are

$x = \frac{a - 3 b \pm \sqrt{{a}^{2} - 6 a b + 5 {b}^{2}}}{2 b}$

The roots will be coincident and real if

${a}^{2} - 6 a b + 5 {b}^{2} = \left(a - 5 b\right) \left(a - b\right) = 0$

or

$a = b$ or $a = 5 b$

Now solving

${x}^{2} + \left(a - b\right) x + \left(a b - {b}^{2} + 1\right) = 0$ we have

$x = \frac{1}{2} \left(- a + b \pm \sqrt{{a}^{2} - 6 a b + 5 {b}^{2} - 4}\right)$

The condition for complex roots is

${a}^{2} - 6 a b + 5 {b}^{2} - 4 < 0$

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

${a}^{2} - 6 a b + 5 {b}^{2} - 4 = - 4 < 0$

Concluding, if $b {x}^{2} - \left(a - 3 b\right) x + b = 0$ has coincident real roots then ${x}^{2} + \left(a - b\right) x + \left(a b - {b}^{2} + 1\right) = 0$ will have complex roots.

Oct 17, 2017

We are given that the equation:

$b {x}^{2} - \left(a - 3 b\right) x + b = 0$

has one real root, therefore the discriminant of this equation is zero:

$\Delta = 0$
$\implies {\left(- \left(a - 3 b\right)\right)}^{2} - 4 \left(b\right) \left(b\right) = 0$
$\therefore {\left(a - 3 b\right)}^{2} - 4 {b}^{2} = 0$
$\therefore {a}^{2} - 6 a b + 9 {b}^{2} - 4 {b}^{2} = 0$
$\therefore {a}^{2} - 6 a b + 5 {b}^{2} = 0$
$\therefore \left(a - 5 b\right) \left(a - b\right) = 0$
$\therefore a = b$, or $a = 5 b$

We seek to show the equation:

${x}^{2} + \left(a - b\right) x + \left(a b - {b}^{2} + 1\right) = 0$

has no real roots. This would require a negative discriminant. The discriminant for this equation is:

$\Delta = {\left(a - b\right)}^{2} - 4 \left(1\right) \left(a b - {b}^{2} + 1\right)$
$\setminus \setminus \setminus = {a}^{2} - 2 a b + {b}^{2} - 4 a b + 4 {b}^{2} - 4$
$\setminus \setminus \setminus = {a}^{2} - 6 a b + 5 {b}^{2} - 4$

And now let us consider the two possible cases that satisfy the first equation:

Case 1: $a = b$

$\Delta = {a}^{2} - 6 a b + 5 {b}^{2} - 4$
$\setminus \setminus \setminus = {\left(b\right)}^{2} - 6 \left(b\right) b + 5 {b}^{2} - 4$
$\setminus \setminus \setminus = {b}^{2} - 6 {b}^{2} + 5 {b}^{2} - 4$
$\setminus \setminus \setminus = - 4$
$\setminus \setminus \setminus < 0$

Case 2: $a = 5 b$

$\Delta = {a}^{2} - 6 a b + 5 {b}^{2} - 4$
$\setminus \setminus \setminus = {\left(5 b\right)}^{2} - 6 \left(5 b\right) b + 5 {b}^{2} - 4$
$\setminus \setminus \setminus = 25 {b}^{2} - 30 {b}^{2} + 5 {b}^{2} - 4$
$\setminus \setminus \setminus = - 4$
$\setminus \setminus \setminus < 0$

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