$(t+1)^{2}+c=0$ $\newline$ In the given equation, $c$ is a constant. The equation has solutions at $t=\frac{3}{2}$ and $t=-\frac{7}{2}$ . What is the value of $c$ ?

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Algebra 2

Quadratic equation with complex roots

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

\newline

In the given equation,

c

is a constant. The equation has solutions at

t=\frac{3}{2}

and

t=-\frac{7}{2}

. What is the value of

c

Substitute $t = \frac{3}{2}$ : Since the solutions to the equation $(t+1)^2 + c = 0$ are $t = \frac{3}{2}$ and $t = -\frac{7}{2}$ , we can substitute each solution into the equation to find the value of $c$ . First, let's substitute $t = \frac{3}{2}$ into the equation. $(t+1)^2 + c = 0$ $\left(\left(\frac{3}{2}\right)+1\right)^2 + c = 0$
Calculate value inside parentheses: Now, calculate the value inside the parentheses. $\newline$ $(\frac{3}{2}) + 1 = \frac{3}{2} + \frac{2}{2} = \frac{5}{2}$ $\newline$ Now, substitute this value back into the equation. $\newline$ $((\frac{5}{2})^{2}) + c = 0$
Isolate $c$ for $t = \frac{3}{2}$ : Next, calculate the square of $\frac{5}{2}$ .
$\left(\frac{5}{2}\right)^2 = \frac{5^2}{2^2} = \frac{25}{4}$
Now, substitute this value back into the equation.
$\frac{25}{4} + c = 0$
Check $t = -\frac{7}{2}$ : To find the value of $c$ , we need to isolate $c$ on one side of the equation. $\newline$ $c = -\left(\frac{25}{4}\right)$ $\newline$ This gives us one possible value for $c$ .
Calculate value inside parentheses: Now, let's check the other solution, $t = -\frac{7}{2}$ , to ensure that it gives us the same value for $c$ . $(t+1)^2 + c = 0$ $\left(-\frac{7}{2}+1\right)^2 + c = 0$
Isolate $c$ for $t = -\frac{7}{2}$ : Calculate the value inside the parentheses. $\newline$ $(-\frac{7}{2}) + 1 = -\frac{7}{2} + \frac{2}{2} = -\frac{5}{2}$ $\newline$ Now, substitute this value back into the equation. $\newline$ $((-\frac{5}{2})^2) + c = 0$
Isolate $c$ for $t = -\frac{7}{2}$ : Calculate the value inside the parentheses. $\newline$ $(-\frac{7}{2}) + 1 = -\frac{7}{2} + \frac{2}{2} = -\frac{5}{2}$ $\newline$ Now, substitute this value back into the equation. $\newline$ $((-\frac{5}{2})^2) + c = 0$ Next, calculate the square of $-\frac{5}{2}$ . $\newline$ $(-\frac{5}{2})^2 = (-5^2)/(2^2) = \frac{25}{4}$ $\newline$ Now, substitute this value back into the equation. $\newline$ $(\frac{25}{4}) + c = 0$
Isolate $c$ for $t = -\frac{7}{2}$ : Calculate the value inside the parentheses. $\newline$ $(-\frac{7}{2}) + 1 = -\frac{7}{2} + \frac{2}{2} = -\frac{5}{2}$ $\newline$ Now, substitute this value back into the equation. $\newline$ $((-\frac{5}{2})^2) + c = 0$ Next, calculate the square of $-\frac{5}{2}$ . $\newline$ $(-\frac{5}{2})^2 = (-5^2)/(2^2) = \frac{25}{4}$ $\newline$ Now, substitute this value back into the equation. $\newline$ $(\frac{25}{4}) + c = 0$ Again, to find the value of $c$ , we need to isolate $c$ on one side of the equation. $\newline$ $c = -(\frac{25}{4})$ $\newline$ This confirms that the value of $c$ is consistent with both solutions.