0=32-50x^(2) What are the solutions to the given equation? Choose 1 answer: (A) x=(4)/(5) (B) x=-(4)/(5) and x=(4)/(5) (c) x=(16)/(25) (D) x=-(16)/(25) and x=(16)/(25)

Given equation and type: Write down the given equation and identify the type of equation. The given equation is 0 = 32 - 50x^2. This is a quadratic equation in the form of ax^2 + bx + c = 0.Rearrange to standard form: Rearrange the equation to standard quadratic form. To solve for x, we need to rearrange the equation to the form ax^2 + bx + c = 0. In this case, we have: 50x^2 = 32Isolate x^2: Divide both sides of the equation by 50 to isolate x^2. x^2 = 32 / 50 x^2 = 0.64Take square root: Take the square root of both sides to solve for x. Since x^2 = 0.64, we take the square root of both sides to find x. Remember that taking the square root of both sides gives us two solutions: one positive and one negative. x = ±√0.64 x = ±0.8Final solutions: Write down the final solutions. The solutions to the equation are x = -0.8 and x = 0.8. These correspond to the choices (B) x = -4/5 and x = 4/5, since 0.8 is equivalent to 4/5.

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0=32-50x^(2)
What are the solutions to the given equation?
Choose 1 answer:
(A)
x=(4)/(5)
(B)
x=-(4)/(5) and
x=(4)/(5)
(c)
x=(16)/(25)
(D)
x=-(16)/(25) and
x=(16)/(25)

$0=32-50 x^{2}$ $\newline$ What are the solutions to the given equation? $\newline$ Choose $1$ answer: $\newline$ (A) $x=\frac{4}{5}$ $\newline$ B $x=-\frac{4}{5}$ and $x=\frac{4}{5}$ $\newline$ (C) $x=\frac{16}{25}$ $\newline$ (D) $x=-\frac{16}{25}$ and $x=\frac{16}{25}$

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Algebra 1

Compare linear and exponential growth

Full solution

0=32-50 x^{2}

\newline

What are the solutions to the given equation?

\newline

Choose

1

answer:

\newline

(A)

x=\frac{4}{5}

\newline

x=-\frac{4}{5}

and

x=\frac{4}{5}

\newline

(C)

x=\frac{16}{25}

\newline

(D)

x=-\frac{16}{25}

and

x=\frac{16}{25}

Given equation and type: Write down the given equation and identify the type of equation. $\newline$ The given equation is $0 = 32 - 50x^2$ . This is a quadratic equation in the form of $ax^2 + bx + c = 0$ .
Rearrange to standard form: Rearrange the equation to standard quadratic form. $\newline$ To solve for $x$ , we need to rearrange the equation to the form $ax^2 + bx + c = 0$ . In this case, we have: $\newline$ $50x^2 = 32$
Isolate $x^2$ : Divide both sides of the equation by $50$ to isolate $x^2$ . $\newline$ $x^2 = \frac{32}{50}$ $\newline$ $x^2 = 0.64$
Take square root: Take the square root of both sides to solve for x. $\newline$ Since $x^2 = 0.64$ , we take the square root of both sides to find $x$ . Remember that taking the square root of both sides gives us two solutions: one positive and one negative. $\newline$ $x = \pm\sqrt{0.64}$ $\newline$ $x = \pm0.8$
Final solutions: Write down the final solutions. $\newline$ The solutions to the equation are $x = -0.8$ and $x = 0.8$ . These correspond to the choices (B) $x = -\frac{4}{5}$ and $x = \frac{4}{5}$ , since $0.8$ is equivalent to $\frac{4}{5}$ .