Complete the table below. The first one has been done for you. For each statement you must state if it is always true, sometimes true or never true, giving a reason in each case. The quadratic equation $ax^2 + bx + c = 0$, $(a \neq 0)$ has 2 real roots. (i) When a real value of $x$ is substituted into $x^2 - 6x + 10$ the result is positive. (ii) If $ax > b$ then $x > \frac{b}{a}$. (iii) The difference between consecutive square numbers is odd.

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Complete the table below - Edexcel - A-Level Maths Pure - Question 7 - 2017 - Paper 2

Question 7

Complete the table below. The first one has been done for you. For each statement you must state if it is always true, sometimes true or never true, giving a reason... show full transcript

Worked Solution & Example Answer:Complete the table below - Edexcel - A-Level Maths Pure - Question 7 - 2017 - Paper 2

Step 1

When a real value of x is substituted into $x^2 - 6x + 10$ the result is positive.

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Answer

To determine if the expression $x^2 - 6x + 10$ is always positive, we can complete the square:

x^2 - 6x + 10 = (x - 3)^2 + 1

Since $(x - 3)^2$ is always non-negative and adding 1 ensures that the expression is always positive for any real $x$ , we conclude that this statement is always true.

Step 2

If ax > b then x > b/a.

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Answer

This statement can be sometimes true. If we assume $a > 0$ , then we can divide both sides of $ax > b$ by $a$ :

x > \frac{b}{a}

However, if $a < 0$ , the inequality reverses, and the conclusion does not hold true. Therefore, the statement is conditionally true based on the value of $a$ .

Step 3

The difference between consecutive square numbers is odd.

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Answer

Let $n$ be a natural number. The consecutive square numbers are $n^2$ and $(n+1)^2$ . The difference between them is given by:

(n+1)^2 - n^2 = 2n + 1

Since $2n$ is always even, adding 1 results in an odd number. Therefore, this statement is always true.

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