If $\displaystyle ax^4 + (2a+b)x^2 + c + a = 0$ has only 2 real roots, which of the following is true? A. b=-2a B. b>-2a C. b<2a D. b<-2a E. none of them above.
Can someone pls explain how to solve this? Thanks in advance~
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If $\displaystyle ax^4 + (2a+b)x^2 + c + a = 0$ has only 2 real roots, which of the following is true? A. b=-2a B. b>-2a C. b<2a D. b<-2a E. none of them above.
Can someone pls explain how to solve this? Thanks in advance~
This is a quadratic equation in $\displaystyle \displaystyle \begin{align*} x^2 \end{align*}$, so let $\displaystyle \displaystyle \begin{align*} X = x^2 \end{align*}$ and the equation becomesQuote:
Originally Posted by shiny718
$\displaystyle \displaystyle \begin{align*} a\,X^2 + \left( 2a + b \right) X + c + a &= 0 \\ X &= \frac{-(2a + b) \pm \sqrt{(2a + b)^2 - 4a(c + a)}}{2a} \\ x^2 &= \frac{-(2a + b) \pm \sqrt{(2a + b)^2 - 4a(c + a)}}{2a} \\ x &= \pm \sqrt{\frac{-(2a + b) \pm \sqrt{(2a + b)^2 - 4a(c + a)}}{2a}} \end{align*}$
This will have exactly two solutions if $\displaystyle \displaystyle \begin{align*} (2a + b)^2 - 4a(c + a) = 0 \end{align*}$ AND $\displaystyle \displaystyle \begin{align*} \frac{-(2a + b)}{2a} > 0 \end{align*}$.
Rearrange the second of these conditions to get your answer.
if $\displaystyle ax^4 + (2a+b)x^2 + c + a = 0$ has but 2 real roots, then $\displaystyle au^2 + (2a+b)u + c + a = 0$ must have but 1 real root, so its discriminant:
$\displaystyle \Delta = (2a+b)^2 - 4(a)(c+a) = 0$
since $\displaystyle (2a+b)^2 - 4(a)(c+a) = 4a^2 + 4ab + b^2 - 4ac - 4a^2 = b^2 + 4(ab - ac)$ we have that:
$\displaystyle b^2 = 4a(c - b)$
EDIT: upon further reflection, and seeing ProveIt's answer, it's clear that although this condition is necessary, it is not sufficient, because u must be non-negative for x to be real.
this means that u = -(2a+b)/2a when Δ = 0. unfortunately, if we do not know if a > 0, we cannot decide if we should have 2a+b > 0, or 2a+b < 0.
one thing is certain, however, if b = -2a, then we only have one root for x.
