Indices, surds and quadratics

Does touching count as a real root?

Yes. A root is an x-value where y = 0. Touching gives one repeated real root; it is not the same as having no real roots.

For x² − 8x + k, completing the square gives (x − 4)² + k − 16. At k = 16 it touches; at k = 17 the minimum is 1 and the curve never reaches zero.

The graph uses x² − 4x + c, whose boundary is c = 4. The practice uses x² − 6x + k. Find its boundary yourself. Different coefficients change the boundary, not the reasoning.

What should I do next?

Try the boundary question in the interactive lesson, then use its feedback to choose a next step. Check whether you found the minimum’s x-coordinate or its height.

Practice records and supported answers are not an ability estimate or a grade prediction.

What you’ll learn

• Simplify rational powers and exact surds.
• Complete the square to find a quadratic’s turning point and solve inequalities.
• Explain how the discriminant counts real roots, including the repeated-root boundary.

Understand the method

A fractional power combines a root and a power: $a^{m/n}=(\sqrt[n]{a})^m$ wherever the real expressions are defined. A negative power gives a reciprocal. Keep surds exact during algebra, and rationalise a denominator by multiplying numerator and denominator by the same suitable expression.

For $ax^2+bx+c$ to be a quadratic, $a\ne0$. Factorisation makes its roots visible; completing the square makes its turning point visible. To complete the square, first factor $a$ out of the two terms containing $x$, halve the coefficient of $x$ inside the bracket, then add and subtract the square of that half. The subtraction keeps the expression equal to the original.

For example, $-x^2+4x-1$ $=-(x^2-4x)-1$ $=-[(x-2)^2-4]-1$ $=-(x-2)^2+3$. Since a square is never negative, the largest value is $3$, reached at $x=2$. In general, $a(x-h)^2+k$ has turning point $(h,k)$: a minimum when $a>0$ and a maximum when $a<0$. Expand your completed square to check every sign and constant.

Why does the discriminant work? Rearranging $ax^2+bx+c=0$ gives $(2ax+b)^2=b^2-4ac$. Write the right-hand side as $\Delta$. If $\Delta>0$, then $2ax+b=\pm\sqrt{\Delta}$, giving two distinct real roots. If $\Delta=0$, both coincide at $x=-b/(2a)$: one repeated root, where the graph touches the axis. If $\Delta<0$, no real square can equal it, so there are no real roots. This argument requires $a\ne0$; if a parameter makes $a=0$, solve the resulting linear or constant equation separately.

For inequalities, locate any real roots, then test the sign in each interval. With two distinct roots, an upward-opening quadratic is negative between them and positive outside; the signs reverse for a downward-opening quadratic. Include a root for $\leq$ or $\geq$, but exclude it for $<$ or $>$. At a repeated root there is no sign change: $(x-2)^2<0$ has no real solutions, whereas $(x-2)^2\leq0$ has only $x=2$. With no real roots the quadratic keeps the sign of $a$ everywhere.

To solve a line and a quadratic simultaneously, substitute the linear expression into the quadratic, solve the resulting equation, then find each matching second coordinate. Check both original equations so that every solution remains a pair.

Key formulae

$$a^m a^n=a^{m+n},\qquad (a^m)^n=a^{mn}$$

$$\begin{aligned}x&=\frac{-b\pm\sqrt{\Delta}}{2a}\\\Delta&=b^2-4ac,\quad a\ne0\end{aligned}$$

$$\begin{aligned}&ax^2+bx+c\\&=a\left(x+\frac{b}{2a}\right)^2+c-\frac{b^2}{4a}\\&\text{where }a\ne0\end{aligned}$$

$$\begin{aligned}\Delta>0&:\ 2\text{ distinct real roots}\\\Delta=0&:\ 1\text{ repeated real root}\\\Delta<0&:\ 0\text{ real roots}\end{aligned}$$

Worked example

Complete the square for $2x^2-12x+10$, find its minimum value, and solve $2x^2-12x+10\leq0$.

1. Factor out $2$ from the terms containing $x$: $2(x^2-6x)+10$. Half of $-6$ is $-3$, so $x^2-6x=(x-3)^2-9$.
2. Substitute and simplify: $2[(x-3)^2-9]+10=2(x-3)^2-8$. The minimum is $-8$ at $x=3$, since $2(x-3)^2\geq0$.
3. For the inequality, $2(x-3)^2-8\leq0$ gives $(x-3)^2\leq4$. Thus $-2\leq x-3\leq2$, so $1\leq x\leq5$.
4. Check: $2x^2-12x+10=2(x-1)(x-5)$ has roots $1$ and $5$. The discriminant is $144-80=64>0$, consistent with two distinct roots; the endpoints belong because the inequality allows equality.

Answer: $2(x-3)^2-8$; minimum $-8$ at $x=3$; inequality solution $1\leq x\leq5$.

Practise quadratic equations and inequalities

Download the free P1 quadratics practice sheet (PDF), then try the questions in this lesson when the interactive page opens.