Question

PLS HELP!! I THIS IS WORTH 200 PTS!! I WILL MARK BRAINIEST WHEN SOMEONE ANSWERS FIRST

Answer 1

14)

Expression: 5z² - 2z + 1 = 0

Discriminant formula: b²- 4ac

Identify: a = 5, b = -2, c = 1

Solve for discriminant: (-2)²- 4(5)(1) = 4 - 20 = -16

15)

• 1st equation: 1m - 1p = 2

p = 1 also can be rewritten as 2nd equation: 0m + 1p = 1

Build Augment Matrices by taking coefficient's:

$\left[\begin{array}{ccc}1&-1&|| \ \ 2\\0&1&|| \ \ 1\end{array}\right]$

Solve:

Add row 2 to row 1 : R₁ = R₁ + R₂ and restore it back

$\rightarrow \ \left[\begin{array}{ccc}1+0&-1+1& \ \ \ \ || \ 2+1\\0&1&|| \ \ 1\end{array}\right]$

$\rightarrow \ \left[\begin{array}{ccc}1&0& || \ \ 3\\0&1&|| \ \ 1\end{array}\right] \ \ \ = The \quad reduced \quad matrix$

Solution = (m, p) = (3, 1)

Answer 2

Answer:

14.  -16

15.  m = 3  and  p = 1

Step-by-step explanation:

Question 14

Discriminant

$b^2-4ac\quad\textsf{when}\quad ax^2+bx+c=0$

Given equation:   $5z^2-2z+1=0$

Therefore:

• a = 5
• b = -2
• c = 1

Substituting the values into the discriminant:

$\begin{aligned}\implies b^2-4ac & = (-2)^2-4(5)(1)\\& = 4-20\\& = -16\end{aligned}$

As the discriminant is negative, this implies there are no real roots.

Question 15

Augmented Matrices

$\begin{array}{c}ax+by=p \\cx+dy=q\end{array} \implies \left[\begin{array}{r r | r}a & b & p \\ c & d & q\end{array}\right]$    

Use elementary row operations to convert into:

$\left[\begin{array}{r r | r}1 & 0 & h \\ 0 & 1 & k\end{array}\right] \implies \begin{array}{r}x=h\\y=k\end{array}$

Given:

$\begin{aligned}m-p & =2\\ p & = 1\end{aligned} \implies \left[\begin{array}{r r | r}1 & -1 & 2 \\ 0 & 1 & 1\end{array}\right]$

$\left[\begin{array}{r r | r}1 & -1 & 2 \\ 0 & 1 & 1\end{array}\right] \begin{array}{c}R_1 + R_2 \rightarrow R_1 \\\rightarrow\end{array} \left[\begin{array}{r r | r}1 & 0 & 3 \\ 0 & 1 & 1\end{array}\right]$

Therefore:  

$m=3$

$p=1$