Q. at2+27t−4=0The 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 equationat2+(27)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.First, let's plug in t=−8.
Substitute t=−8: Substitute t=−8 into the equation:a(−8)2+(27)(−8)−4=0Simplify the equation:64a−28−4=064a−32=0
Solve for a (t=−8): Now, solve for a:64a=32a=6432a=21
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.Substitute t=1 into the original equation:a(1)2+(27)(1)−4=0Simplify the equation:a+27−4=0
Solve for a (t=1): Now, solve for a using the equation from the previous step:a=4−27a=28−27a=21
Verify Consistency: The value of a is consistent for both solutions t=−8 and t=1. Therefore, the value of the constant a is 21.
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