Answer:
[tex]\textsf{1. \quad $k = -10, \;\; k = 10$}[/tex]
[tex]\textsf{2. \quad $k < \dfrac{16}{3}$}[/tex]
[tex]\textsf{3. \quad $k < \dfrac{4}{7}$}[/tex]
Step-by-step explanation:
[tex]\boxed{\begin{minipage}{7.6 cm}\underline{Discriminant}\\\\$b^2-4ac$ \quad when $ax^2+bx+c=0$\\\\when $b^2-4ac > 0 \implies$ two real solutions.\\when $b^2-4ac=0 \implies$ one real solution.\\when $b^2-4ac < 0 \implies$ no real solutions.\\\end{minipage}}[/tex]
Question 1
Given equation:
[tex]5x^2 + kx + 5 = 0[/tex]
Therefore:
[tex]a=5, \quad b=k, \quad c=5[/tex]
If the equation has one real solution, set the discriminant equal to zero and solve for k:
[tex]\begin{aligned}b^2-4ac&=0\\\implies k^2-4(5)(5)&=0\\k^2-100&=0\\k^2&=100\\\sqrt{k^2}&=\sqrt{100}\\k&= \pm 10\end{aligned}[/tex]
Question 2
Given equation:
[tex]3x^2 + 8x + k = 0[/tex]
Therefore:
[tex]a=3, \quad b=8, \quad c=k[/tex]
If the equation has two real solutions, set the discriminant to greater than zero and solve for k:
[tex]\begin{aligned}b^2-4ac& > 0\\\implies 8^2-4(3)(k)& > 0\\64-12k& > 0\\64& > 12k\\12k& < 64\\k& < \dfrac{16}{3}\end{aligned}[/tex]
Question 3
Given equation:
[tex]kx^2 - 4x + 7 =0[/tex]
Therefore:
[tex]a=k, \quad b=-4, \quad c=7[/tex]
If the equation has two real solutions, set the discriminant to greater than zero and solve for k:
[tex]\begin{aligned}b^2-4ac& > 0\\\implies (-4)^2-4(k)(7)& > 0\\16-28k& > 0\\16& > 28k\\28k& < 16\\k& < \dfrac{16}{28}\\k& < \dfrac{4}{7}\end{aligned}[/tex]
