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NeX [460]
2 years ago
7

Determine a quadratic equation, in standard form, that has the pair of root: x=-3, x=5

Mathematics
2 answers:
vlabodo [156]2 years ago
6 0

Answer:

x²-2x-15=0

Step-by-step explanation:

x=-3 x=5

x+3=0 x-5=0

(x+3)(x-5)=0

x²-5x+3x-15=0

x²-2x-15=0

Mama L [17]2 years ago
5 0

Answer:

x^2 -2x -15=0

Step-by-step explanation:

\text{Given that, the roots are,}~ \alpha=-3 ~ \text{and}~ \beta=5 .\\\\\text{The quadratic equation with roots}~ \alpha ~ \text{and}~ \beta ~ \text{is,}\\\\~~~~~~~~x^2 - (\alpha + \beta )x +\alpha \beta = 0\\\\\implies x^2-(-3 +5) x +(-3)(5) = 0\\\\\implies x^2 -2x -15=0

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3 years ago
Review the attachments. Review the process of solving an equation and fill in the blanks.
Nezavi [6.7K]

Answer:

1) Combine like terms

2) \sqrt[3]{x} =3

3) cube both sides of the equation

4) 4\sqrt[3]{27} +8\sqrt[3]{27}=36

5) 4(3) + 8(3) = 36

Step-by-step explanation:

1) Combine like terms

2) \sqrt[3]{x} =3

3) cube both sides of the equation

4) 4\sqrt[3]{27} +8\sqrt[3]{27}=36

5) 4(3) + 8(3) = 36

5 0
2 years ago
Use Cramer Rule to solve the following system: 8x−5y=70 and 9x+7y=3
nlexa [21]

Answer:

(x,y) = (5,-6)

Step-by-step explanation:

\underline{\textbf{Determinant of a matrix.}}\\\\\text{For a}~ 2 \times 2 ~ \text{matrix,}\\\\\begin{vmatrix} a_1&a_2\\b_1&b_2 \end{vmatrix} = a_1b_2 - a_2b_1\\\\\\\text{For a}~ 3 \times 3 ~ \text{matrix,}\\\\\begin{vmatrix} a_1&a_2&a_3\\ b_1&b_2&b_3\\ c_1&c_2&c_3 \end{vmatrix} = a_1\begin{vmatrix} b_2&b_3\\c_2&c_3 \end{vmatrix} - a_2 \begin{vmatrix} b_1&b_3\\c_1&c_3 \end{vmatrix}+ a_3 \begin{vmatrix} b_1&b_2\\c_1&c_2 \end{vmatrix}\\\\\\

                     ~~~~~~~~~~~~~~~~~~=a_1(b_2c_3-b_3c_2) -a_2(b_1c_3-b_3c_1) +a_3(b_1c_2-b_2c_1)

\underline{\textbf{Cramer's Rule to solve a system of two equations.}}\\\\\text{Consider the system of two equations:}\\\\~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~a_1x + b_1 y= c_1\\\\~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~a_2x +b_2 y = c_2\\\\\text{Here,}\\\\x = \dfrac{D_x}{D}= \dfrac{\begin{vmatrix} c_1&b_1\\c_2&b_2 \end{vmatrix}}{\begin{vmatrix} a_1&b_1\\a_2&b_2 \end{vmatrix}}\\\\\\ y= \dfrac{D_y}{D}= \dfrac{\begin{vmatrix} a_1&c_1\\a_2&c_2 \end{vmatrix}}{\begin{vmatrix} a_1&b_1\\a_2&b_2 \end{vmatrix}}\\\\

\underline{\textbf{Solution:}}\\\\~~~~~~~~~~~~~~~~~~~~~~~8x-5y = 70~~~~~~...(i)\\\\~~~~~~~~~~~~~~~~~~~~~~~9x +7y = 3~~~~~~~...(ii)\\\\\text{Applying Cramer's rule:}\\\\x = \dfrac{D_x}{D}\\\\\\~~=\dfrac{\begin{vmatrix} 70& -5 \\3&7 \end{vmatrix}}{\begin{vmatrix} 8& -5\\ 9& 7\end{vmatrix}}\\\\\\~~=\dfrac{70(7) -(-5)(3)}{(8)(7)-(-5)(9)}\\\\\\~~=\dfrac{490+15}{56+45}\\\\\\~~=\dfrac{505}{101}\\\\\\~~=5

y = \dfrac{D_y}{D}\\\\\\~~=\dfrac{\begin{vmatrix} 8& 70 \\9&3 \end{vmatrix}}{\begin{vmatrix} 8& -5\\ 9& 7\end{vmatrix}}\\\\\\~~=\dfrac{(8)(3) -(70)(9)}{(8)(7)-(-5)(9)}\\\\\\~~=\dfrac{24-630}{56+45}\\\\\\~~=-\dfrac{606}{101}\\\\\\~~=-6

\textbf{Hence, the solution to the system of equation is}~ (x,y) = (5,-6)

7 0
2 years ago
NEED HELP ASAP, WORTH 80PTS
sp2606 [1]

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This is because the expression can be simplified to get the same result. -1+2log_4((1/4)x)

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