(t+1)^(2)+c=0 In the given equation, c is a constant. The equation has solutions at t=(3)/(2) and t=-(7)/(2). What is the value of c ? Choose 1 answer: (A) -(729)/(4) (B) -(121)/(4) (C) -(25)/(4) (D) -1

Step 1: Substituting t = 3/2: The given equation is in the form of a quadratic equation, and we are told that t = 3/2 and t = -7/2 are solutions to this equation. This means that if we substitute these values of t into the equation, the equation should hold true. Let's start by substituting t = 3/2 into the equation.Step 2: Solving for c: Substitute t = 3/2 into the equation (t+1)² + c = 0: ((3/2) + 1)² + c = 0 (5/2)² + c = 0 (25/4) + c = 0Step 3: Verifying with t = -7/2: Now, we need to solve for c by moving (25/4) to the other side of the equation: c = -(25/4)Step 4: Confirming the value of c: Next, let's verify our result by substituting the other solution, t = -7/2, into the original equation to ensure that it also satisfies the equation with the same value of c.Substitute t = -7/2 into the equation (t+1)² + c = 0: ((-7/2) + 1)² + c = 0 (-5/2)² + c = 0 (25/4) + c = 0Since we have already found that c = -(25/4), substituting this value into the equation should result in a true statement: (25/4) - (25/4) = 0 0 = 0 This confirms that our value for c is correct.

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Math Problems

Algebra 1

Transformations of absolute value functions: translations and reflections

Full solution

(t+1)^{2}+c=0

\newline

In the given equation,

c

is a constant.

\newline

The equation has solutions at

t=\frac{3}{2}

and

t=-\frac{7}{2}

. What is the value of

c

\newline

Choose

1

answer:

\newline

(A)

-\frac{729}{4}

\newline

(B)

-\frac{121}{4}

\newline

(C)

-\frac{25}{4}

\newline

(D)

-1

Step $1$ : Substituting $t = \frac{3}{2}$ : The given equation is in the form of a quadratic equation, and we are told that $t = \frac{3}{2}$ and $t = -\frac{7}{2}$ are solutions to this equation. This means that if we substitute these values of $t$ into the equation, the equation should hold true. Let's start by substituting $t = \frac{3}{2}$ into the equation.
Step $2$ : Solving for c: Substitute $t = \frac{3}{2}$ into the equation $(t+1)^2 + c = 0$ : $\newline$ $\left(\frac{3}{2} + 1\right)^2 + c = 0$ $\newline$ $\left(\frac{5}{2}\right)^2 + c = 0$ $\newline$ $\frac{25}{4} + c = 0$
Step $3$ : Verifying with $t = -\frac{7}{2}$ : Now, we need to solve for $c$ by moving $\frac{25}{4}$ to the other side of the equation: $\newline$ $c = -\left(\frac{25}{4}\right)$
Step $4$ : Confirming the value of $c$ : Next, let's verify our result by substituting the other solution, $t = -\frac{7}{2}$ , into the original equation to ensure that it also satisfies the equation with the same value of $c$ .
Step $4$ : Confirming the value of c: Next, let's verify our result by substituting the other solution, $t = -\frac{7}{2}$ , into the original equation to ensure that it also satisfies the equation with the same value of $c$ .Substitute $t = -\frac{7}{2}$ into the equation $(t+1)^2 + c = 0$ : $\newline$ $\left(-\frac{7}{2} + 1\right)^2 + c = 0$ $\newline$ $\left(-\frac{5}{2}\right)^2 + c = 0$ $\newline$ $\left(\frac{25}{4}\right) + c = 0$
Step $4$ : Confirming the value of c: Next, let's verify our result by substituting the other solution, $t = -\frac{7}{2}$ , into the original equation to ensure that it also satisfies the equation with the same value of c.Substitute $t = -\frac{7}{2}$ into the equation $(t+1)^2 + c = 0$ : $\newline$ $\left(-\frac{7}{2} + 1\right)^2 + c = 0$ $\newline$ $\left(-\frac{5}{2}\right)^2 + c = 0$ $\newline$ $\left(\frac{25}{4}\right) + c = 0$ Since we have already found that $c = -\left(\frac{25}{4}\right)$ , substituting this value into the equation should result in a true statement: $\newline$ $\left(\frac{25}{4}\right) - \left(\frac{25}{4}\right) = 0$ $\newline$ $0 = 0$ $\newline$ This confirms that our value for c is correct.