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(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?

(t+1)2+c=0 (t+1)^{2}+c=0 \newlineIn the given equation, c c is a constant. The equation has solutions at t=32 t=\frac{3}{2} and t=−72 t=-\frac{7}{2} . What is the value of c c ?

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Q. (t+1)2+c=0 (t+1)^{2}+c=0 \newlineIn the given equation, c c is a constant. The equation has solutions at t=32 t=\frac{3}{2} and t=−72 t=-\frac{7}{2} . What is the value of c c ?
  1. Substitute t=32t = \frac{3}{2}: Since the solutions to the equation (t+1)2+c=0(t+1)^2 + c = 0 are t=32t = \frac{3}{2} and t=−72t = -\frac{7}{2}, we can substitute each solution into the equation to find the value of cc. First, let's substitute t=32t = \frac{3}{2} into the equation. (t+1)2+c=0(t+1)^2 + c = 0 ((32)+1)2+c=0\left(\left(\frac{3}{2}\right)+1\right)^2 + c = 0
  2. Calculate value inside parentheses: Now, calculate the value inside the parentheses.\newline(32)+1=32+22=52(\frac{3}{2}) + 1 = \frac{3}{2} + \frac{2}{2} = \frac{5}{2}\newlineNow, substitute this value back into the equation.\newline((52)2)+c=0((\frac{5}{2})^{2}) + c = 0
  3. Isolate cc for t=32t = \frac{3}{2}: Next, calculate the square of 52\frac{5}{2}.
    (52)2=5222=254\left(\frac{5}{2}\right)^2 = \frac{5^2}{2^2} = \frac{25}{4}
    Now, substitute this value back into the equation.
    254+c=0\frac{25}{4} + c = 0
  4. Check t=−72t = -\frac{7}{2}: To find the value of cc, we need to isolate cc on one side of the equation.\newlinec=−(254)c = -\left(\frac{25}{4}\right)\newlineThis gives us one possible value for cc.
  5. Calculate value inside parentheses: Now, let's check the other solution, t=−72t = -\frac{7}{2}, to ensure that it gives us the same value for cc.(t+1)2+c=0(t+1)^2 + c = 0(−72+1)2+c=0\left(-\frac{7}{2}+1\right)^2 + c = 0
  6. Isolate cc for t=−72t = -\frac{7}{2}: Calculate the value inside the parentheses.\newline(−72)+1=−72+22=−52(-\frac{7}{2}) + 1 = -\frac{7}{2} + \frac{2}{2} = -\frac{5}{2}\newlineNow, substitute this value back into the equation.\newline((−52)2)+c=0((-\frac{5}{2})^2) + c = 0
  7. Isolate cc for t=−72t = -\frac{7}{2}: Calculate the value inside the parentheses.\newline(−72)+1=−72+22=−52(-\frac{7}{2}) + 1 = -\frac{7}{2} + \frac{2}{2} = -\frac{5}{2}\newlineNow, substitute this value back into the equation.\newline((−52)2)+c=0((-\frac{5}{2})^2) + c = 0Next, calculate the square of −52-\frac{5}{2}.\newline(−52)2=(−52)/(22)=254(-\frac{5}{2})^2 = (-5^2)/(2^2) = \frac{25}{4}\newlineNow, substitute this value back into the equation.\newline(254)+c=0(\frac{25}{4}) + c = 0
  8. Isolate cc for t=−72t = -\frac{7}{2}: Calculate the value inside the parentheses.\newline(−72)+1=−72+22=−52(-\frac{7}{2}) + 1 = -\frac{7}{2} + \frac{2}{2} = -\frac{5}{2}\newlineNow, substitute this value back into the equation.\newline((−52)2)+c=0((-\frac{5}{2})^2) + c = 0Next, calculate the square of −52-\frac{5}{2}.\newline(−52)2=(−52)/(22)=254(-\frac{5}{2})^2 = (-5^2)/(2^2) = \frac{25}{4}\newlineNow, substitute this value back into the equation.\newline(254)+c=0(\frac{25}{4}) + c = 0Again, to find the value of cc, we need to isolate cc on one side of the equation.\newlinec=−(254)c = -(\frac{25}{4})\newlineThis confirms that the value of cc is consistent with both solutions.

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