Final answer:
The question involves solving for x in quadratic and linear equations, considering real-world applications where negative results may not be applicable. Decimals and scientific notation conversion also form a part of the problem-solving process.
Step-by-step explanation:
The student is working on solving quadratic equations and linear equations, and the goal is to find the value of x.
When using the quadratic equation, the student has found two potential solutions for x, which are x = -0.0024 and x = 0.00139. However, given the context, negative solutions are not applicable, so the real-life value for x is 0.00139.
When given a scientific notation like x x 10¹ where x is 4.5, to convert it to a standard number, we need to move the decimal point. Here, since y is negative, the decimal point moves to the left six places, turning 4.5 x 10⁶ into 0.0000045.
Further simplifying equations when x is significantly smaller than other terms, allows us to approximate (0.25 - x) as 0.25 which simplifies the solving process. Ultimately, in another example, when evaluating a quadratic equation, the student finds x equals 0.0216 or -0.0224.