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

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

Solve the equation $x^2 - 2x - 4 = 0$. Give your answers in the form $a \pm \sqrt{b}$, where $a, b \in \mathbb{N}$. Given that $\left(\sqrt{d}\right)^2 = d$, mult... 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 equation x2−2x−4=0x^2 - 2x - 4 = 0, we can apply the quadratic formula:
x=−b±b2−4ac2ax = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
In this case, a=1a = 1, b=−2b = -2, and c=−4c = -4.
Calculating the discriminant:
b2−4ac=(−2)2−4(1)(−4)=4+16=20b^2 - 4ac = (-2)^2 - 4(1)(-4) = 4 + 16 = 20
Therefore,
x=2±202x = \frac{2 \pm \sqrt{20}}{2}
Simplifying this gives:
x=1±5x = 1 \pm \sqrt{5}

Step 2

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

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Answer

Using the identity (a+b)2=a2+2ab+b2(a + b)^2 = a^2 + 2ab + b^2, we can expand:
(c+d)2=c2+2cd+d\left(c + \sqrt{d}\right)^2 = c^2 + 2c\sqrt{d} + d

Step 3

Complete the table

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Answer

To complete the table:

  • For 6\sqrt{6}: Yes for Q\mathbb{Q} and No for R∖Q\mathbb{R} \setminus \mathbb{Q} as it is irrational.
  • For 2: Yes for all (N,Z,Q\mathbb{N}, \mathbb{Z}, \mathbb{Q}), No for R∖Q\mathbb{R} \setminus \mathbb{Q}.
  • For 3: Yes for all (N,Z,Q\mathbb{N}, \mathbb{Z}, \mathbb{Q}), No for R∖Q\mathbb{R} \setminus \mathbb{Q}.
  • For -4: No for N\mathbb{N}, Yes for Z\mathbb{Z} and Q\mathbb{Q}, No for R∖Q\mathbb{R} \setminus \mathbb{Q}.
    This gives the completed table:

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