# The nature of roots of the quadratic equation $ax^2+(b-c)x-2b-c-a=0,$

If the expression $ax^2+2bx+c$, where $a$ is a non-zero real number, has the same sign as that of $a$ for every real value of $x$, then roots of the quadratic equation $ax^2+(b-c)x-2b-c-a=0$ are:

(A) real and equal
(B) real and unequal
(C) non-real having positive real part
(D) non-real having having negative real part

As the expression $ax^2+2bx+c$ has the same sign as that of $a$ for every real value of $x$, so if $a>0,$ then $4b^2-4ac<0$ and if $a<0$, then $4b^2-4ac>0$

To determine the nature of roots of the equation $ax^2+(b-c)x-2b-c-a=0, I found its discriminant$\Delta =(b-c)^2+4a(2b+c+a)=b^2+c^2-2bc+8ab+4ac+4a^2$Now I am not able to find the nature of roots of the equation. • I think you mean "discriminant," not "determinant" – ASKASK Feb 1 '16 at 3:53 • I corrected it. – Brahmagupta Feb 1 '16 at 3:55 • I am sorry @DavidK ,i corrected it. – Brahmagupta Feb 1 '16 at 4:07 • Since$ax^2+2bx+c$has always the same sign as$a$for any real$x$, it has no real roots, so$4b^2 - 4ac < 0$. That is true no matter whether$a$is positive or negative. If$4b^2 - 4ac > 0$then you have two real roots and$ax^2+2bx+c$takes on both positive and negative values, contradicting the given. – David K Feb 1 '16 at 4:09 • Yes,that is true point,but how to decide the nature of roots of the given equation. – Brahmagupta Feb 1 '16 at 4:11 ## 1 Answer Since$ax^2+2bx+c$has always the same sign as$a$for any real$x$, it has no real roots, so$4b^2 - 4ac < 0\$.

Now try writing \begin{align} (b-c)^2 + 4a(2b+c+a) &= (b-c)^2 + 4(2ab+ac+a^2) \\ &= (b-c)^2 + 4(2ab+b^2+a^2) + 4(ac - b^2) . \end{align}

• That means,discriminant of the given equation is strictly greater than zero.@DavidK,Am i correct? – Brahmagupta Feb 1 '16 at 4:18
• Yes, you can show it is the sum of two squares and a positive expression. – David K Feb 1 '16 at 4:26