Algebra: Quadratics

Question

$$4x^2+(5k+3)x+(2k^2-1)=0$$

find the value of $k$ for which the roots are numerically the same except opposite in sign I tried substituting a and -a and use quadratic formula but it comes up with a variable solution and I was looking for a solid number solution.

Answer 1

For $$ax^2+bx+c=0,$$ where $a\neq0$ we need $b=0$ and $\frac{c}{a}<0$.

Thus, $k=-\frac{3}{5}$ and check that $2k^2-1<0$.

Answer 2

$4x^2+(5k+3)x+(2k^2-1)=4x^2-\alpha^2$

$k=\frac{-3}5$

${ \alpha} ^2=1-2\times \frac{9}{25}=\frac{7}{25}$

$ \alpha =\pm \frac{ \sqrt {7}}{5}$

$\alpha \in \mathbb{R}$ so value of $k$ is valid.

Answer 3

$$ Q(x) = 4x^2 + (5k+3)x + \left(2k^2-1\right) = 0 \\ Q(x) = Ax^2 + Bx + C = 0 \\ $$

For the zeroes of the quadratic $Q(x)$ to be the same in magnitude but opposite in sign, then $Q(x)$ must be symmetrical about the axis $x=0$. The key here is to know that the axis of symmetry of a parabola is $x=-b/(2a)$.

This implies that

$$\begin{align} \frac{-B}{2A} &= 0 \\ -B &= 0 \\ B &= 0 \\ 5k+3 &= 0 \\ 5k &= -3 \\ k &= -\frac35 \\ \end{align}$$