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Answer: (5 - √1) and (4 + √7)
Step-by-step explanation:
Here we have 4 roots, then this is a polynomial of 4th degree.
We can write it as
a*x⁴ + b*x³ + c*x² + d*x + e = 0
We could rewrite this as the product of two quadratic equations:
(a₁*x² + b₁*x + c₁)*(a₂*x² + b₂*x + c₂) = 0
This will be equal to zero when (a₁*x² + b₁*x + c₁) is equal to zero, and when
(a₂*x² + b₂*x + c₂) is equal to zero.
Then we need to find the roots of these two quadratic equations.
And we know that the roots of a quadratic equation like:
a₁*x² + b₁*x + c₁ are given by:
Then the general form of the roots is something like:
This means that if one of the roots is:
5 + √1, we must have another root equal to 5 - √1
And if we have one root equal to 4 - √7, then we must have another root equal to 4 + √7
Then the correct options are:
(5 - √1) and (4 + √7)
❄ Hi there,
, rounded to 3 DP (decimal places)
Why round? Because the entire number goes on for ever and ever.
So, we need to round. 1/6 rounded to 3 decimal places is 0.167. We have 3 numbers after the decimal point.
❄