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Solve using the quadratic formula. $\newline$ $w^2 + 4w - 5 = 0$ $\newline$ Write your answers as integers, proper or improper fractions in simplest form, or decimals rounded to the nearest hundredth. $\newline$ $w =$ _ or $w =$ _

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Solve a quadratic equation using the quadratic formula

Full solution

Q. Solve using the quadratic formula.

\newline

w^2 + 4w - 5 = 0

\newline

Write your answers as integers, proper or improper fractions in simplest form, or decimals rounded to the nearest hundredth.

\newline

w =

_____ or

w =

_____

Identify values of $a$ , $b$ , $c$ : Identify the values of $a$ , $b$ , and $c$ in the quadratic equation $w^2 + 4w - 5 = 0$ . The standard form of a quadratic equation is $ax^2 + bx + c = 0$ . Comparing this with our equation, we get: $a = 1$ $b = 4$ $b$ $0$
Substitute into quadratic formula: Substitute the values of $a$ , $b$ , and $c$ into the quadratic formula $w = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ . Substitute $a = 1$ , $b = 4$ , and $c = -5$ into the quadratic formula. $w = \frac{-(4) \pm \sqrt{(4)^2 - 4\cdot1\cdot(-5)}}{2\cdot1}$
Calculate discriminant: Simplify the expression under the square root and calculate the discriminant. $\newline$ Calculate $\sqrt{(4)^2 - 4\cdot 1\cdot (-5)}$ . $\newline$ $\sqrt{16 + 20}$ $\newline$ $\sqrt{36}$
Simplify square root: Simplify the square root and continue with the formula. $\newline$ Since $\sqrt{36}$ is $6$ , we have: $\newline$ $w = (\frac{-4 \pm 6}{2})$
Calculate solutions for w: Calculate the two possible solutions for w. $\newline$ First solution: $\newline$ w = $(-4 + 6) / 2$ $\newline$ w = $2 / 2$ $\newline$ w = $1$ $\newline$ Second solution: $\newline$ w = $(-4 - 6) / 2$ $\newline$ w = $-10 / 2$ $\newline$ w = $-5$

More problems from Solve a quadratic equation using the quadratic formula

Question

h

is defined over the real numbers. This table gives a few values of

h

\newline

\begin{tabular}{lllll}

\newline

x

-6

1

-6

01

-6

001

-5

9

\newline

h(x)

-0

25

-0

74

-0

98

-1

0

\newline

\newline

What is a reasonable estimate for

\lim _{x \rightarrow-6} h(x)

\newline

1

\newline

-6

\newline

-2

\newline

-1

\newline

(D) The limit doesn't exist

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\newline

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