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at^(2)+(7)/(2)t-4=0
The given equation has solutions at
t=-8 and
t=1. What is the value of the constant
a ?

$a t^{2}+\frac{7}{2} t-4=0$ $\newline$ The given equation has solutions at $t=-8$ and $t=1$ . What is the value of the constant $a$ ?

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Write and solve direct variation equations

Full solution

Q.

a t^{2}+\frac{7}{2} t-4=0

\newline

The given equation has solutions at

t=-8

and

t=1

. What is the value of the constant

a

?

Given Quadratic Equation: We are given that the quadratic equation $a t^{2} + \left(\frac{7}{2}\right)t - 4 = 0$ has solutions $t = -8$ and $t = 1$ . We can use these solutions to find the value of the constant $a$ by plugging them into the equation and solving for $a$ . $\newline$ First, let's plug in $t = -8$ .
Substitute $t = -8$ : Substitute $t = -8$ into the equation: $\newline$ $a(-8)^2 + \left(\frac{7}{2}\right)(-8) - 4 = 0$ $\newline$ Simplify the equation: $\newline$ $64a - 28 - 4 = 0$ $\newline$ $64a - 32 = 0$
Solve for $a$ ( $t = -8$ ): Now, solve for $a$ : $64a = 32$ $a = \frac{32}{64}$ $a = \frac{1}{2}$
Substitute $t = 1$ : We have found a value for $a$ using the solution $t = -8$ . However, we must also verify that this value of $a$ works for the other solution, $t = 1$ , to ensure that there is no math error. $\newline$ Substitute $t = 1$ into the original equation: $\newline$ $a(1)^2 + (\frac{7}{2})(1) - 4 = 0$ $\newline$ Simplify the equation: $\newline$ $a + \frac{7}{2} - 4 = 0$
Solve for $a$ ( $t = 1$ ): Now, solve for $a$ using the equation from the previous step: $\newline$ $a = 4 - \frac{7}{2}$ $\newline$ $a = \frac{8}{2} - \frac{7}{2}$ $\newline$ $a = \frac{1}{2}$
Verify Consistency: The value of $a$ is consistent for both solutions $t = -8$ and $t = 1$ . Therefore, the value of the constant $a$ is $\frac{1}{2}$ .

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