If 4x^(2)-12 x=40, what are the possible values for x ? A) -2 and 5 B) -5 and 2 C) -5 and -8 D) -5 and -16

Rewrite in Standard Form: Start by rewriting the given equation in standard quadratic form. The given equation is 4x^2 - 12x = 40. To rewrite it in standard form, we need to set the equation equal to zero. 4x^2 - 12x - 40 = 0Factor the Quadratic Equation: Factor the quadratic equation. To factor the equation, we look for two numbers that multiply to give -160 (4 * -40) and add to give -12 (the coefficient of x). The two numbers that satisfy these conditions are -20 and +8. So we can write the equation as: 4x^2 - 20x + 8x - 40 = 0 Grouping the terms, we get: (4x^2 - 20x) + (8x - 40) = 0 Taking out the common factors, we have: 4x(x - 5) + 8(x - 5) = 0 Now we can factor out (x - 5): (4x + 8)(x - 5) = 0Solve for x: Solve for x. We have the factored form (4x + 8)(x - 5) = 0. Setting each factor equal to zero gives us the possible values for x: 4x + 8 = 0 or x - 5 = 0 Solving each equation for x gives us: 4x = -8 or x = 5 x = -2 or x = 5Verify Solutions: Verify the solutions. We substitute x = -2 and x = 5 back into the original equation to check if they are valid solutions. For x = -2: 4(-2)^2 - 12(-2) = 40 16 + 24 = 40 40 = 40 (True) For x = 5: 4(5)^2 - 12(5) = 40 100 - 60 = 40 40 = 40 (True) Both solutions satisfy the original equation.

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If
4x^(2)-12 x=40, what are the possible values for
x ?
A) -2 and 5
B) -5 and 2
C) -5 and -8
D) -5 and -16

If $4 x^{2}-12 x=40$ , what are the possible values for $x$ ? $\newline$ A) $-2$ and $5$ $\newline$ B) $-5$ and $2$ $\newline$ C) $-5$ and $-8$ $\newline$ D) $-5$ and $-16$

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Algebra 2

Solve complex trigonomentric equations

Full solution

Q. If

4 x^{2}-12 x=40

, what are the possible values for

x

\newline

-2

and

5

\newline

-5

and

2

\newline

-5

and

-8

\newline

-5

and

-16

Rewrite in Standard Form: Start by rewriting the given equation in standard quadratic form. $\newline$ The given equation is $4x^2 - 12x = 40$ . To rewrite it in standard form, we need to set the equation equal to zero. $\newline$ $4x^2 - 12x - 40 = 0$
Factor the Quadratic Equation: Factor the quadratic equation. $\newline$ To factor the equation, we look for two numbers that multiply to give $-160$ $(4 \times -40)$ and add to give $-12$ (the coefficient of $x$ ). $\newline$ The two numbers that satisfy these conditions are $-20$ and $+8$ . $\newline$ So we can write the equation as: $\newline$ $4x^2 - 20x + 8x - 40 = 0$ $\newline$ Grouping the terms, we get: $\newline$ $(4x^2 - 20x) + (8x - 40) = 0$ $\newline$ Taking out the common factors, we have: $\newline$ $4x(x - 5) + 8(x - 5) = 0$ $\newline$ Now we can factor out $(x - 5)$ : $\newline$ $(4 \times -40)$ $0$
Solve for x: Solve for x. $\newline$ We have the factored form $(4x + 8)(x - 5) = 0$ . $\newline$ Setting each factor equal to zero gives us the possible values for x: $\newline$ $4x + 8 = 0$ or $x - 5 = 0$ $\newline$ Solving each equation for x gives us: $\newline$ $4x = -8$ or $x = 5$ $\newline$ $x = -2$ or $x = 5$
Verify Solutions: Verify the solutions. $\newline$ We substitute $x = -2$ and $x = 5$ back into the original equation to check if they are valid solutions. $\newline$ For $x = -2$ : $\newline$ $4(-2)^2 - 12(-2) = 40$ $\newline$ $16 + 24 = 40$ $\newline$ $40 = 40$ (True) $\newline$ For $x = 5$ : $\newline$ $4(5)^2 - 12(5) = 40$ $\newline$ $100 - 60 = 40$ $\newline$ $40 = 40$ (True) $\newline$ Both solutions satisfy the original equation.