Quadratic Equations - Solution of a Quadratic Equation by Factorisation

Find the roots of the quadratic equation $6x^2 - x - 2 = 0$ by factorisation.

Given: $6x^2 - x - 2 = 0$ Here, $a = 6$, $b = -1$, and $c = -2$. We need to find $p$ and $q$ such that: $p + q = -1$ $p \times q = 6 \times (-2) = -12$ The numbers are $-4$ and $3$. Split the middle term: $6x^2 - 4x + 3x - 2 = 0$ $2x(3x - 2) + 1(3x - 2) = 0$ $(3x - 2)(2x + 1) = 0$ Now, either $3x - 2 = 0$ or $2x + 1 = 0$. $3x = 2 \implies x = \frac{2}{3}$ $2x = -1 \implies x = -\frac{1}{2}$ Roots are $\frac{2}{3}$ and $-\frac{1}{2}$.

We identify $ac = -12$ and $b = -1$. We choose $-4$ and $3$ because their sum is $-1$ and product is $-12$. Then we group the terms to factorise.

Solve for $x$: $\sqrt{2}x^2 + 7x + 5\sqrt{2} = 0$.

We find $p, q$ such that $p + q = 7$ and $pq = (\sqrt{2})(5\sqrt{2})$. Calculating $ac$: $\begin{array}{r} \sqrt{2} \times \sqrt{2} = 2 \\ 2 \times 5 = 10 \\ \hline ac = 10 \end{array}$ We need $p + q = 7$ and $pq = 10$. The numbers are $5$ and $2$. $\sqrt{2}x^2 + 5x + 2x + 5\sqrt{2} = 0$ $x(\sqrt{2}x + 5) + \sqrt{2}(\sqrt{2}x + 5) = 0$ $(\sqrt{2}x + 5)(x + \sqrt{2}) = 0$ Equating factors to zero: $\sqrt{2}x + 5 = 0 \implies x = -\frac{5}{\sqrt{2}}$ $x + \sqrt{2} = 0 \implies x = -\sqrt{2}$

Even with irrational coefficients, the splitting method remains the same. Note that $2$ can be written as $\sqrt{2} \times \sqrt{2}$ to facilitate factorisation by grouping.