Solve the equation $x^2 - 2x - 4 = 0$. Give your answers in the form $\alpha \pm \sqrt{\beta}$, where $\alpha, \beta \in \mathbb{N}$. Given that $\sqrt{a}^2 = d$, multiply out and simplify $\left(c + \sqrt{a}\right)^2$. The table below shows whether each of the given numbers is an element of the natural numbers ($\mathbb{N}$), the integers ($\mathbb{Z}$), the rational numbers ($\mathbb{Q}$), and the irrational numbers ($\mathbb{R} \setminus \mathbb{Q}$). Complete the table by writing "Yes" or "No" into each box. One row is already done for you: 16 is an element of $\mathbb{N}$, $\mathbb{Z}$, and $\mathbb{Q}$, but not of $\mathbb{R} \setminus \mathbb{Q}$.

Junior Cycle Mathematics ALGEBRA - Factorising expressions: Solve the equation $x^2 - 2x

Solve the equation $x^2 - 2x - 4 = 0$ - Junior Cycle Mathematics - Question 9 - 2017

Question 9

Solve the equation $x^2 - 2x - 4 = 0$. Give your answers in the form $\alpha \pm \sqrt{\beta}$, where $\alpha, \beta \in \mathbb{N}$. Given that $\sqrt{a}^2 = d$, m... show full transcript

Worked Solution & Example Answer:Solve the equation $x^2 - 2x - 4 = 0$ - Junior Cycle Mathematics - Question 9 - 2017

Step 1

Solve the equation $x^2 - 2x - 4 = 0$

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Answer

To solve the quadratic equation $x^2 - 2x - 4 = 0$ , we can use the quadratic formula:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

In this case, $a = 1$ , $b = -2$ , and $c = -4$ .

Calculating the discriminant:

$b^2 - 4ac = (-2)^2 - 4(1)(-4) = 4 + 16 = 20$

Now substituting the values into the formula:

$x = \frac{-(-2) \pm \sqrt{20}}{2(1)}$

This simplifies to:

$x = \frac{2 \pm \sqrt{20}}{2}$

Further simplification gives:

$x = 1 \pm \sqrt{5}$

Thus, the solutions are $x = 1 + \sqrt{5}$ and $x = 1 - \sqrt{5}$ .

Step 2

Given that $\sqrt{a}^2 = d$, multiply out and simplify $(c + \sqrt{a})^2$

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Answer

To simplify $(c + \sqrt{a})^2$ , we use the expansion for the square of a binomial:

$(c + \sqrt{a})^2 = c^2 + 2c \sqrt{a} + \left(\sqrt{a}\right)^2$

Since $\left(\sqrt{a}\right)^2 = a$ , we can rewrite the expression as:

$c^2 + 2c \sqrt{a} + a$

Step 3

Complete the table with the given numbers

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Answer

Number	$\mathbb{N}$	$\mathbb{Z}$	$\mathbb{Q}$	$\mathbb{R} \setminus \mathbb{Q}$
16	Yes	Yes	Yes	No
$\sqrt{6}$	No	No	Yes	Yes
2	Yes	Yes	Yes	No
-4	No	Yes	Yes	No

Solve the equation $x^2 - 2x - 4 = 0$ - Junior Cycle Mathematics - Question 9 - 2017

Worked Solution & Example Answer:Solve the equation $x^2 - 2x - 4 = 0$ - Junior Cycle Mathematics - Question 9 - 2017

Solve the equation $x^2 - 2x - 4 = 0$

Given that $\sqrt{a}^2 = d$, multiply out and simplify $(c + \sqrt{a})^2$

Complete the table with the given numbers

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