The area of triangle ABC is $\sqrt{2}$ m$^2$ - Edexcel - GCSE Maths - Question 15 - 2017 - Paper 3

We know from the question that the area is $\sqrt{2}$ m$^2$. Therefore, we set up the equation:

$\frac{1}{2} \times (x+3) \times (2x-1) = \sqrt{2}$

Multiplying both sides by 2 gives:

$(x+3)(2x-1) = 2\sqrt{2}$

Expanding the left side:

$2x^2 + 6x - x - 3 = 2\sqrt{2}$

This simplifies to:

$2x^2 + 5x - 3 = 2\sqrt{2}$

Rearranging the equation gives:

$2x^2 + 5x - (2\sqrt{2} + 3) = 0$

The solutions for $x$ can now be found using the quadratic formula:

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

Here, $a = 2$, $b = 5$, and $c = -(2\sqrt{2} + 3)$. Substituting these values into the formula:

$x = \frac{-5 \pm \sqrt{5^2 - 4 \times 2 \times (-(2\sqrt{2} + 3))}}{2 \times 2}$

After substituting and simplifying, we solve for $x$. Finally, calculate the numerical values to obtain:

$x \approx 0.414, -3.414$

As $x$ must be a positive length, the valid solution is:

$x \approx 0.414$

Thus, rounding to 3 significant figures, we find:

$x = 0.414$