Step-by-step solution of an Oxford access exam problem
Every year, thousands of students face the challenge of the admission exams to enter the prestigious University of Oxford, a process that tests not only their knowledge, but also their ability to solve under pressure. In particular, the math exam is known for its complexity, posing problems of all kinds, from solving equations to questions of logic, advanced algebra, calculus and number theory. In this article, we will explore one of the problems presented in October 2023.
The problem poses the following question: how many real solutions does the following equation have?
Do you dare to solve the equation before reading the complete solution in the blog?
At first glance, the statement seems simple: an equation with a single variable, where all the numbers involved are integers between 1 and 4. However, the real difficulty lies in correctly undoing the squared parentheses, making sure to consider all possible cases.
A key detail in this type of equations is that, when squaring a number, the same result is obtained for both its positive and negative values. For example:
This occurs because squaring a number eliminates the negative sign. Therefore, when solving a quadratic equation we must take into account the two possible options: the positive and the negative number that can generate that result.
One last observation: if we were to fully develop the parentheses, we would obtain an equation of the form xˆ16+…, which indicates that the equation is of degree 16 and, at most, could have 16 real solutions for x. Therefore, we cannot rule out any of the options presented to us as possible answers.
Since for equations of degree 16 there is no systematic formula like Ruffini’s method for polynomials of degree 3, we will solve this equation by working the parentheses progressively, from the outermost to the innermost.
Specific example of how to solve an equation of degree 16
Let’s get on it!
Let’s start with the outermost parenthesis. If we define t as ((x²-1)²-2)²-3 , then we get t²=4. This gives us two possible values for t: t=2 or t=-2.
Substituting t by its original value in each case, we obtain the following results:
1. If t=2 then: ((x²-1)²-2)²-3 =2 ⇒ ((x²-1)²-2)²=5 and again, substituting r= (x²-1)²-2 we obtain that r²=5 . Let’s look at the two new cases that arise:
1.1 r=√5: substituting r for its original value we obtain (x²-1)²-2 =√5 ⇒ (x²-1)²=2+√5 and again, one last time, we substitute s=x²-1, we obtain that s²=2+√5 and observe the two results again:
When trying to calculate the square root to obtain the value of x, we would be taking the root of a negative number. As a result, the two values obtained will be imaginary numbers.
1.2 r=-√5 substituting r for its original value we obtain (x²-1)²-2=-√5 ⇒ (x²-1)²=2-√5 . Following the same reasoning as in the previous section, given that √4=2, it follows that √5>2 . This implies that 2-√5<0, and when trying to calculate its square root, we would obtain an imaginary number. Therefore, when we continue developing to obtain the value of x, this would also be an imaginary number.
2. If t=-2 then: ((x²-1)²-2)²-3 =-2 ⇒ ((x²-1)²-2)²=1 and again, substituting r= (x²-1)²-2 we obtain that r²=1 . Let’s look at the two cases:
2.1 r=1: substituting r for its original value we obtain (x²-1)²-2 =1 ⇒ (x²-1)²=3 and again, one last time, we substitute s=x²-1, we obtain that s²=3 and observe the two results again:
2.1.1 s=√3 and substituting s for its value, we finally obtain: x²-1=√3 ⇒ x²=1+√3 and therefore
2.1.2 s=-√3 substituting s for its original value we obtain x²-1=-√3 ⇒ x²=1-√3 . This subtraction is negative, since √2≈1,41 and therefore, √3>1,41 with which we deduce that 1-√3<0 , and when trying to calculate its square root, we would obtain an imaginary number.
2.2 r=-1 substituting r for its original value we obtain (x²-1)²-2 = -1 ⇒ (x²-1)² = 1 and again, one last time, we substitute s=x²-1 and we obtain that s²=1 and observe the two results again:
2.2.1 s=1 and substituting s for its value, we finally obtain: x²-1 = 1 ⇒ x²=2 and therefore
2.2.1.1 x=√2 which is a real solution.
2.2.1.2 x=-√2 which is a real solution.
2.2.2 s=-1 and substituting s for its value, we finally obtain: x²-1 = -1 ⇒ x²=0 and therefore the only possible solution is x=0 .
Let’s review all the possible real values we have obtained:
This gives us a total of 7 real solutions for x, so the correct answer is option (c).
Did you get your answer right?
Drawing this equation with our graphing tool is a simple and effective way to check the number of real solutions. Looking at the graph, you can clearly see how the curve interacts with the x-axis, confirming the previous result. Below, we show you the graph to see for yourself:
Done with the WirisQuizzes Assessment tool
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