Question

Find the value(s) of k that will cause the equation to have the given number and type of solutions.

1. 5x^2 + kx + 5 = 0, 1 real solution

2. 3x^2 + 8x + k = 0, 2 real solutions

3. kx^2 - 4x + 7 =0, 2 real solutions

Fastest one who solves the problems gets Brainliest; thanks in advance! <3

Answer 1

Answer:

$\textsf{1. \quad k = -10, \;\; k = 10 }$

$\textsf{2. \quad k < \dfrac{16}{3} }$

$\textsf{3. \quad k < \dfrac{4}{7} }$

Step-by-step explanation:

$\boxed{\begin{minipage}{7.6 cm}\underline{Discriminant}\\\\ b^2-4ac \quad when ax^2+bx+c=0 \\\\when b^2-4ac > 0 \implies two real solutions.\\when b^2-4ac=0 \implies one real solution.\\when b^2-4ac < 0 \implies no real solutions.\\\end{minipage}}$

Question 1

Given equation:

$5x^2 + kx + 5 = 0$

Therefore:

$a=5, \quad b=k, \quad c=5$

If the equation has one real solution, set the discriminant equal to zero and solve for k:

$\begin{aligned}b^2-4ac&=0\\\implies k^2-4(5)(5)&=0\\k^2-100&=0\\k^2&=100\\\sqrt{k^2}&=\sqrt{100}\\k&= \pm 10\end{aligned}$

Question 2

Given equation:

$3x^2 + 8x + k = 0$

Therefore:

$a=3, \quad b=8, \quad c=k$

If the equation has two real solutions, set the discriminant to greater than zero and solve for k:

$\begin{aligned}b^2-4ac& > 0\\\implies 8^2-4(3)(k)& > 0\\64-12k& > 0\\64& > 12k\\12k& < 64\\k& < \dfrac{16}{3}\end{aligned}$

Question 3

Given equation:

$kx^2 - 4x + 7 =0$

Therefore:

$a=k, \quad b=-4, \quad c=7$

If the equation has two real solutions, set the discriminant to greater than zero and solve for k:

$\begin{aligned}b^2-4ac& > 0\\\implies (-4)^2-4(k)(7)& > 0\\16-28k& > 0\\16& > 28k\\28k& < 16\\k& < \dfrac{16}{28}\\k& < \dfrac{4}{7}\end{aligned}$