All categories

3 years ago

Find the exact solution of the following quadratic equation by using the Quadratic Formula. x^2-9x=36

Mathematics

1 answer:

natita [175]3 years ago

4 0

Answer:

x = 12 or x = -3

Step-by-step explanation:

Solve for x over the real numbers:

x^2 - 9 x = 36

Subtract 36 from both sides:

x^2 - 9 x - 36 = 0

x = (9 ± sqrt((-9)^2 - 4 (-36)))/2 = (9 ± sqrt(81 + 144))/2 = (9 ± sqrt(225))/2:

x = (9 + sqrt(225))/2 or x = (9 - sqrt(225))/2

sqrt(225) = sqrt(9×25) = sqrt(3^2×5^2) = 3×5 = 15:

x = (9 + 15)/2 or x = (9 - 15)/2

(9 + 15)/2 = 24/2 = 12:

x = 12 or x = (9 - 15)/2

(9 - 15)/2 = -6/2 = -3:

Answer: x = 12 or x = -3

You might be interested in

Which expression is equal to 4/8

givi [52]

<h2>Fractions</h2>

Which expression is equal to $\large\tt \frac{4}{8}$ ?

Option 1

$\large \tt 4 \times \frac{1}{2} = \frac{4}{1} \times \frac{1}{2} = \red{\frac{4}{2}}$

Option 1 is not equal to $\large\tt \frac{4}{8}$

$\huge{\overline{\qquad\qquad\qquad\qquad \: \: \: }}$

Option 2

$\large \tt 3 \times \frac{1}{5} = \frac{3}{1} \times \frac{1}{5} = \red{\frac{3}{5}}$

Option 2 is not equal to $\large\tt \frac{4}{8}$

$\huge{\overline{\qquad\qquad\qquad\qquad \: \: \: }}$

Option 3

$\large \tt 3 \times \frac{1}{8} = \frac{3}{1} \times \frac{1}{8} = \red{\frac{3}{8}}$

Option 3 is not equal to $\large\tt \frac{4}{8}$

$\huge{\overline{\qquad\qquad\qquad\qquad \: \: \: }}$

Option 4

$\large \tt 4 \times \frac{1}{8} = \frac{4}{1} \times \frac{1}{8} = \green{\frac{4}{8}}$

Therefore, option 4 is equal to $\large\tt \frac{4}{8}$

$\huge{\overline{\qquad\qquad\qquad\qquad \: \: \: }}$

<h3>#CarryOnLearning</h3>

7 0

3 years ago

Which coordinate pair fits the set of coordinates {(0, 6), (-2, 14 3 ), (3, 8)} defined by a linear function?

sattari [20]

19,14

$(19.4$

4 0

3 years ago

Find the? inverse, if it? exists, for the given matrix.<br><br> [4 3]<br><br> [3 6]

True [87]

Answer:

Therefore, the inverse of given matrix is

$=\begin{pmatrix}\frac{2}{5}&-\frac{1}{5}\\ -\frac{1}{5}&\frac{4}{15}\end{pmatrix}$

Step-by-step explanation:

The inverse of a square matrix $A$ is $A^{-1}$ such that

$A A^{-1}=I$ where I is the identity matrix.

Consider, $A = \left[\begin{array}{ccc}4&3\\3&6\end{array}\right]$

$\mathrm{Matrix\:can\:only\:be\:inverted\:if\:it\:is\:non-singular,\:that\:is:}$

$\det \begin{pmatrix}4&3 \\3&6\end{pmatrix}\ne 0$

$\mathrm{Find\:2x2\:matrix\:inverse\:according\:to\:the\:formula}:\quad \begin{pmatrix}a\:&\:b\:\\ c\:&\:d\:\end{pmatrix}^{-1}=\frac{1}{\det \begin{pmatrix}a\:&\:b\:\\ c\:&\:d\:\end{pmatrix}}\begin{pmatrix}d\:&\:-b\:\\ -c\:&\:a\:\end{pmatrix}$

$=\frac{1}{\det \begin{pmatrix}4&3\\ 3&6\end{pmatrix}}\begin{pmatrix}6&-3\\ -3&4\end{pmatrix}$

$\mathrm{Find\:the\:matrix\:determinant\:according\:to\:formula}:\quad \det \begin{pmatrix}a\:&\:b\:\\ c\:&\:d\:\end{pmatrix}\:=\:ad-bc$

$4\cdot \:6-3\cdot \:3=15$

$=\frac{1}{15}\begin{pmatrix}6&-3\\ -3&4\end{pmatrix}$

$=\begin{pmatrix}\frac{2}{5}&-\frac{1}{5}\\ -\frac{1}{5}&\frac{4}{15}\end{pmatrix}$

Therefore, the inverse of given matrix is

$=\begin{pmatrix}\frac{2}{5}&-\frac{1}{5}\\ -\frac{1}{5}&\frac{4}{15}\end{pmatrix}$

4 0

3 years ago

X varies directly with<br> y. If x = 60 when<br> y = 75. find x when<br> y = 32.

Alex Ar [27]

Answer:

x=17

Step-by-step explanation:

3 0

3 years ago

Read 2 more answers

Deandre wants to earn at least $30 trimming trees. He charges $7 per hour and pays $5 in equipment fees. What are the possible n

exis [7]

8 hours for the possible numbers of house

5 0

3 years ago