Consider the equation $4x^2 − mx − m = 0 $ (with unknown x).
Find the values of m such that the equation has
(a) one (double) root;
(b) two distinct (real) roots;
(c) no real roots.
My working out
$4x^2 − mx − m = 0 $
Apply the quadratic formula gives
$m+/-\sqrt{m^2+4m\:-16} / 2(4)$
a) $m^2 +4m − 16 = 0 $
Through completing the square I get:
$(m + 2)^2 -20 = 0$
$m = -2$
b)$m^2 + 4m − 16 > 0$
$(m + 2)^2 -20 > 0$
$m+2> 0$
I can't seem to find more solutions
c) $m^2 + 4m − 16 < 0$
HINT: Consider the discriminant $D=b^2 - 4ac$. Then $ax^2 + bx + c = 0$ has two different real solution iff $D>0$, double root iff $D=0$ and complex roots iff $D<0$
The solutions of inequalities are wrong. Try to solve $m^2 + 4m - 16=0$ and write $m^2 + 4m - 16 = (m-m_1)(m-m_2)$, where $m_1,m_2$ are the solutions of the equation. Then we have that:
$$m^2 + 4m - 16 >0 \iff (m-m_1)> 0 \text{ and } (m-m_2)>0 \text{ or } (m-m_1)< 0 \text{ and } (m-m_2)<0$$
$$m^2 + 4m - 16 =0 \iff (m-m_1)= 0 \text{ or } (m-m_2)=0$$
$$m^2 + 4m - 16 <0 \iff (m-m_1)< 0 \text{ and } (m-m_2)>0 \text{ or } (m-m_1)> 0 \text{ and } (m-m_2)<0$$
$$0 = m^2 + 4m - 16 = (m+2)^2 - 20 \iff (m+2)^2 = 20$$ $$\iff m+2 = \pm \sqrt{20} \iff m = -2 \pm \sqrt{20}$$
– Stefan4024 Nov 02 '16 at 21:07