A-Level AQA Maths Pure Proof: Given the quadratic expressions

Question

Given the quadratic expressions $x^2 + bx + c$ and $x^2 + dx + e$, we know that they have a common factor, specifically $(x + 2)$. To show that $2(d - b) = e - c$, we can use the factor theorem, which states that if $(x + 2)$ is a factor, then substituting $x = -2$ should yield zero for the respective expressions.

First, substituting $x = -2$ into the first expression gives:

$$(-2)^2 + b(-2) + c = 0 \ 4 - 2b + c = 0$$

From this, we can rearrange to find:

$$c = 2b - 4 $$

Next, substituting $x = -2$ into the second expression yields:

$$(-2)^2 + d(-2) + e = 0 \ 4 - 2d + e = 0 $$

Rearranging gives:

$$e = 2d - 4$$

Now we have two equations:

1. $c = 2b - 4$
2. $e = 2d - 4$

To find the relationship between $2(d - b)$ and $e - c$, we can substitute our expressions for $c$ and $e$:

$$e - c = (2d - 4) - (2b - 4)$$
$$= 2d - 2b$$
$$= 2(d - b)$$

Thus, we have shown that:

$$2(d - b) = e - c$$

This completes the proof.

Given the quadratic expressions $x^2 + bx + c$ and $x^2 + dx + e$, we know that they have a common factor, specifically $(x + 2)$ - AQA - A-Level Maths Pure - Question 4 - 2019 - Paper 2

Worked Solution & Example Answer:Given the quadratic expressions $x^2 + bx + c$ and $x^2 + dx + e$, we know that they have a common factor, specifically $(x + 2)$ - AQA - A-Level Maths Pure - Question 4 - 2019 - Paper 2

Show that $2(d - b) = e - c$

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