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3 years ago

The quadratic equation is

Mathematics

1 answer:

NemiM [27]3 years ago

8 0

Answer:

(2,-9)

Step-by-step explanation:

Set the quadratic equal to zero.

$y = {x}^{2} - 4x - 5$

Factor

$(x - 5)(x + 1)$

Set each of these factors to zero

$x - 5 = 0$

and

$x + 1 = 0$

$x = 5$

and

$x = - 1$

Plot the midpoint of the two values.

$( \frac{5 + ( - 1)}{2} ) = 2$

So the middle term is 2 and the other terms are 5 and -1.

Now the vertex is

$2 {}^{2} - 4(2) - 5 = - 9$

So the vertex is

(2,-9)

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2 years ago

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Read 2 more answers

Solving a Two-Step Matrix Equation Solve the equation:

Cloud [144]

Answer:

$\boxed {x_{1} = 3}$

$\boxed {x_{2} = -4}$

Step-by-step explanation:

Solve the following equation:

$\left[\begin{array}{ccc}3&2\\5&5\\\end{array}\right] \left[\begin{array}{ccc}x_{1}\\x_{2}\\\end{array}\right] + \left[\begin{array}{ccc}1\\2\\\end{array}\right] = \left[\begin{array}{ccc}2\\-3\\\end{array}\right]$

-In order to solve a pair of equations by using substitution, you first need to solve one of the equations for one of variables and then you would substitute the result for that variable in the other equation:

-First equation:

$3x_{1} + 2x_{2} + 1 = 2$

-Second equation:

$5x_{1} + 5x_{2} + 2 = -3$

-Choose one of the two following equations, which I choose the first one, then you solve for $x_{1}$ by isolating

$3x_{1} + 2x_{2} + 1 = 2$

-Subtract $1$ to both sides:

$3x_{1} + 2x_{2} + 1 - 1 = 2 - 1$

$3x_{1} + 2x_{2} = 1$

-Subtract $2x_{2}$ to both sides:

$3x_{1} + 2x_{2} - 2x_{2} = -2x_{2} + 1$

$3x_{1} = -2x_{2} + 1$

-Divide both sides by $3$ :

$3x_{1} = -2x_{2} + 1$

$x_{1} = \frac{1}{3} (-2x_{2} + 1)$

-Multiply $-2x_{2} + 1$ by $\frac{1}{3}$ :

$x_{1} = \frac{1}{3} (-2x_{2} + 1)$

$x_{1} = -\frac{2}{3}x_{2} + \frac{1}{3}$

-Substitute $-\frac{2x_{2} + 1}{3}$ for $x_{1}$ in the second equation, which is $5x_{1} + 5x_{2} + 2 = -3$ :

$5x_{1} + 5x_{2} + 2 = -3$

$5(-\frac{2}{3}x_{2} + \frac{1}{3}) + 5x_{2} + 2 = -3$

Multiply $-\frac{2x_{2} + 1}{3}$ by $5$ :

$5(-\frac{2}{3}x_{2} + \frac{1}{3}) + 5x_{2} + 2 = -3$

$-\frac{10}{3}x_{2} + \frac{5}{3} + 5x_{2} + 2 = -3$

-Combine like terms:

$-\frac{10}{3}x_{2} + \frac{5}{3} + 5x_{2} + 2 = -3$

$\frac{5}{3}x_{2} + \frac{11}{3} = -3$

-Subtract $\frac{11}{3}$ to both sides:

$\frac{5}{3}x_{2} + \frac{11}{3} - \frac{11}{3} = -3 - \frac{11}{3}$

$\frac{5}{3}x_{2} = -\frac{20}{3}$

-Multiply both sides by $\frac{5}{3}$ :

$\frac{\frac{5}{3}x_{2}}{\frac{5}{3}} = \frac{-\frac{20}{3}}{\frac{5}{3}}$

$\boxed {x_{2} = -4}$

-After you have the value of $x_2$ , substitute for $x_{2}$ onto this equation, which is $x_{1} = -\frac{2}{3}x_{2} + \frac{1}{3}$ :

$x_{1} = -\frac{2}{3}x_{2} + \frac{1}{3}$

$x_{1} = -\frac{2}{3}(-4) + \frac{1}{3}$

-Multiply $-\frac{2}{3}$ and $-4$ :

$x_{1} = -\frac{2}{3}(-4) + \frac{1}{3}$

$x_{1} = \frac{8 + 1}{3}$

-Since both $\frac{1}{3}$ and $\frac{8}{3}$ have the same denominator, then add the numerators together. Also, after you have added both numerators together, reduce the fraction to the lowest term:

$x_{1} = \frac{8 + 1}{3}$

$x_{1} = \frac{9}{3}$

$\boxed {x_{1} = 3}$

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3 years ago