Estimate square root of 90 to the nearest hundredths place
2 answers:
The method for doing this will depend on the required accuracy of your estimate.
Using linear approximation between 9^2 = 81 and 10^2 = 100, the estimate is
.. 9 +(90-81)/(100-81) = 9 +9/19 ≈ 9.47
Using one round of "Babylonian Method" on the above estimate,
.. (9.47 +90/9.47)/2 ≈ 9.486 ≈ 9.49 . . . . . . actually accurate to the hundredths place
Using derivatives and linear approximation,
.. √90 ≈ √81 +1/(2√81)*(90-81) = 9.50
or
.. √90 ≈ √100 +1/(2√100)*(90 -100) = 9.50
Using linear approximation between 9^2 = 81 and 10^2 = 100, the estimate is
.. 9 +(90-81)/(100-81) = 9 +9/19 ≈ 9.47
Using one round of "Babylonian Method" on the above estimate,
.. (9.47 +90/9.47)/2 ≈ 9.486 ≈ 9.49 . . . . . . actually accurate to the hundredths place
Using derivatives and linear approximation,
.. √90 ≈ √81 +1/(2√81)*(90-81) = 9.50
or
.. √90 ≈ √100 +1/(2√100)*(90 -100) = 9.50
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Answer:
Step-by-step explanation:
A quadratic equation can be factorised if and only if there are rational roots.
For any quadratic equation the discriminant decides about the nature of roots.
Thus only if the discriminant is a perfect square we can have rational roots and in this case only factorization is possible.
In the given equation
Since 1 is a perfect square we can factor and solve