# If the roots of the equation ax^2+2bx+c=0 are real and distinct then find the nature of the roots of the equation (a+c)(ax^2+2bx+c) = 2(ac - b^2)(x^2+1)?

Jun 19, 2016

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

has a Complex conjugate pair of non-Real zeros.

#### Explanation:

Discriminant

The discriminant $\Delta$ of a quadratic expression $A {x}^{2} + B x + C$ is ${B}^{2} - 4 A C$.

The sign of the discriminant determines the kind of zeros the quadratic has:

• $\Delta > 0$ : Two distinct Real zeros.
• $\Delta = 0$ : One repeated Real zero.
• $\Delta < 0$ : A Complex conjugate pair of non-Real zeros.

Solution

The discriminant of $a {x}^{2} + 2 b x + c$ is:

${\Delta}_{1} = {\left(2 b\right)}^{2} - 4 a c = 4 \left({b}^{2} - a c\right)$

Since $a {x}^{2} + 2 b x + c = 0$ has distinct Real zeros, ${\Delta}_{1} > 0$, hence:

${b}^{2} - a c > 0$

Let $k = {b}^{2} - a c > 0$

Given:

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

Expand and rearrange into standard form:

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

This has discriminant:

${\Delta}_{2} = {\left(2 b \left(a + c\right)\right)}^{2} - 4 \left({a}^{2} - a c + 2 {b}^{2}\right) \left({c}^{2} - a c + 2 {b}^{2}\right)$

$= 4 \left({b}^{2} {\left(a + c\right)}^{2} - \left({a}^{2} - a c + 2 {b}^{2}\right) \left({c}^{2} - a c + 2 {b}^{2}\right)\right)$

$= 4 \left(\left(a c + k\right) {\left(a + c\right)}^{2} - \left({a}^{2} + a c + 2 k\right) \left({c}^{2} + a c + 2 k\right)\right)$

$= 4 \left(\textcolor{red}{\cancel{\textcolor{b l a c k}{a c {\left(a + c\right)}^{2}}}} + k {\left(a + c\right)}^{2} - \textcolor{red}{\cancel{\textcolor{b l a c k}{a c {\left(a + c\right)}^{2}}}} - 2 k {\left(a + c\right)}^{2} - 4 {k}^{2}\right)$

$= 4 \left(- k {\left(a + c\right)}^{2} - 4 {k}^{2}\right) < 0$

So:

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

has a Complex conjugate pair of non-Real zeros.