Find algebraically the set of values of $x$ for which
$x^3 - 49 > 0$ and $5x^3 - 31x - 72 > 0$. - Edexcel - GCSE Maths - Question 1 - 2022 - Paper 3
Question 1
Find algebraically the set of values of $x$ for which
$x^3 - 49 > 0$ and $5x^3 - 31x - 72 > 0$.
Worked Solution & Example Answer:Find algebraically the set of values of $x$ for which
$x^3 - 49 > 0$ and $5x^3 - 31x - 72 > 0$. - Edexcel - GCSE Maths - Question 1 - 2022 - Paper 3
Step 1
$x^3 - 49 > 0$
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Answer
To solve the inequality x3−49>0, we first find the roots by setting the equation equal to zero:
x3−49=0
This can be factored as:
(x−7)(x2+7x+49)=0
The real root is x=7. Now consider the behavior of the function:
For x<7, x3−49<0.
For x=7, x3−49=0.
For x>7, x3−49>0.
Thus, the solution to x3−49>0 is:
x>7
Step 2
$5x^3 - 31x - 72 > 0$
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Answer
To solve the inequality 5x3−31x−72>0, we first find the roots by setting:
5x3−31x−72=0
Using methods such as synthetic division or the Rational Root Theorem, we can test possible rational roots. By testing, we find:
x=3 is a root.
Now factor the polynomial:
5x3−31x−72=(x−3)(5x2+15x+24)
The quadratic 5x2+15x+24 has a negative discriminant, indicating it has no real roots.
Thus, we can analyze the polynomial:
For x<3, 5x3−31x−72<0.
For x=3, 5x3−31x−72=0.
For x>3, 5x3−31x−72>0.
Hence, the solution to 5x3−31x−72>0 is:
x>3
Step 3
Combining the solutions
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Answer
To find the values of x that satisfy both inequalities:
From the first part, we have x>7.
From the second part, we have x>3.
Since x>7 is a stricter constraint than x>3, the combined solution set is:
x>7
