2.4 Solving quadratic equations, including in a function of the unknown — Practise with help

1. $x^2 + 7x + 12 = (x + 3)(x + 4)$.
2. $b^2 - 4ac = 2^2 - 4 \times 1 \times 5 = 4 - 20 = -16$. It is negative, so there are no real roots.
3. $2^3 = 8$ and $2^{-1} = \frac{1}{2}$.

If question 1 was hard, revisit the factor pairs of $12$: you need a pair that adds to $7$.

Look for two numbers that multiply to $10$ and add to $-7$.

Write the equation as $(x - 5)(x - 2) = 0$, then set each factor to zero.

Factorise:

So $x - 5 = 0$ or $x - 2 = 0$, giving $x = 5$ or $x = 2$.

The roots are not whole numbers, so complete the square. Half of $6$ is $3$.

Write $x^2 + 6x + 2 = (x + 3)^2 - 7$, set it to zero, and rearrange to $(x + 3)^2 = 7$.

Complete the square:

So $(x + 3)^2 = 7$, and $x + 3 = \pm\sqrt{7}$. Therefore

Identify $a = 3$, $b = -5$ and $c = -2$, then substitute into $x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$.

Work out the discriminant first: $(-5)^2 - 4 \times 3 \times (-2) = 25 + 24 = 49$.

Substitute into the formula:

So $x = \frac{12}{6} = 2$ or $x = \frac{-2}{6} = -\frac{1}{3}$. Check by factorising: $3x^2 - 5x - 2 = (3x + 1)(x - 2)$.

The expression is a quadratic in $x^2$. Let $t = x^2$ and solve for $t$.

With $t = x^2$ the equation becomes $t^2 - 10t + 9 = 0$, which factorises as $(t - 1)(t - 9) = 0$.

Let $t = x^2$. Then

so $t = 1$ or $t = 9$. Replace $t$ with $x^2$:

Taking both signs gives $x = \pm 1$ and $x = \pm 3$, so there are four solutions: $x = 1$, $x = -1$, $x = 3$ and $x = -3$.

Here $3^{2x} = \left(3^x\right)^2$, so let $t = 3^x$.

Solve $t^2 - 10t + 9 = 0$, then replace $t$ with $3^x$ and match both sides to powers of $3$.

Let $t = 3^x$. Then

so $t = 1$ or $t = 9$. Replace $t$:

Since $3^0 = 1$ and $3^2 = 9$, the solutions are $x = 0$ and $x = 2$.

Square both sides to remove the root. Squaring can create a value that does not work, so every answer must be checked.

Squaring gives $2x + 3 = x^2$, so rearrange to $x^2 - 2x - 3 = 0$ and factorise.

Square both sides:

The two values are $x = 3$ and $x = -1$. Check the original equation. For $x = 3$: $\sqrt{2 \times 3 + 3} = \sqrt{9} = 3$, which matches the right side. For $x = -1$: $\sqrt{1} = 1$, but the right side is $-1$, so $x = -1$ fails. The only solution is $x = 3$.