GCSE Edexcel Maths Expanding Brackets: Find algebraically the set of va

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 2

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 2

Step 1

Solve $x^3 - 49 > 0$

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Answer

To solve the inequality $x^3 - 49 > 0$ , we can first identify where the equation equals zero:

$x^3 - 49 = 0$

This can be factored as: $x^3 = 49$ Thus, we have: $x = 7$ Next, we need to test intervals around this critical point to determine where the inequality holds.

For $x < 7$ , let’s test $x = 0$ :
$0^3 - 49 = -49 < 0$ (not in solution region)
For $x > 7$ , let’s test $x = 8$ :
$8^3 - 49 = 512 - 49 = 463 > 0$ (in solution region)

Therefore, the solution for this part is: $x > 7$

Step 2

Solve $5x^3 - 31x - 72 > 0$

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Answer

To solve the inequality $5x^3 - 31x - 72 > 0$ , we first find the roots of the corresponding equation:

$5x^3 - 31x - 72 = 0$

Using the Rational Root Theorem or numerical methods, we find that one root is $x = 3$ .

Next, we factor the cubic polynomial or use synthetic division to check for additional roots:

$5x^3 - 31x - 72 = (x - 3)(5x^2 + 15x + 24)$ The quadratic $5x^2 + 15x + 24$ has no real roots (as the discriminant $15^2 - 4(5)(24) < 0$ ).

Now we analyze the sign of $5x^2 + 15x + 24$ :

For all real values of $x$ , the quadratic is always positive because its leading coefficient (5) is positive. Therefore, the inequality $5x^3 - 31x - 72 > 0$ holds when: $x > 3$

Combining both parts, the overall solution for the given question is: $x > 7$ and $x > 3$ . So, the final solution is just: $x > 7$

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 2

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 2

Solve $x^3 - 49 > 0$

Solve $5x^3 - 31x - 72 > 0$

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