Quadratic Functions, Worked Examples (easy)

The word 'solve' asks for the values of $x$ that make the equation true, and the coefficients here are small whole numbers, so factorising is the quickest route. We need two numbers that multiply to $-15$ (the constant term) and add to $+2$ (the coefficient of $x$).

List the factor pairs of $-15$: $(1,-15)$, $(-1,15)$, $(3,-5)$ and $(-3,5)$. The pair $+5$ and $-3$ works, since $5\times(-3)=-15$ and $5+(-3)=2$.

So the equation factorises as:

A product equals zero only when one of the factors is zero, so set each bracket to zero in turn:

The roots are $x=-5$ and $x=3$. It is worth checking at least one: substituting $x=3$ gives $(3)^2+2(3)-15 = 9+6-15 = 0$, and substituting $x=-5$ gives $(-5)^2+2(-5)-15 = 25-10-15 = 0$, so both roots are confirmed.

The formula works for every quadratic, so it is the safe choice when the coefficient of $x^2$ is not $1$ and factors are harder to spot. The quadratic formula is $x = \dfrac{-b \pm \sqrt{b^2-4ac}}{2a}$.

Read off the coefficients carefully: here $a=3$, $b=5$ and $c=-2$, keeping the sign of $c$.

First evaluate the discriminant $b^2-4ac$, taking care with the double negative:

Since $49$ is a perfect square, $\sqrt{49}=7$, which tells us the roots will be rational. Substitute into the formula, remembering that the whole numerator sits over $2a$:

Take the two signs separately to get the two roots:

So $x=\tfrac{1}{3}$ or $x=-2$. Substituting $x=-2$ gives $3(-2)^2+5(-2)-2 = 12-10-2 = 0$, and $x=\tfrac13$ gives $3(\tfrac19)+5(\tfrac13)-2 = \tfrac13+\tfrac53-2 = 0$, so both roots check out.

Here the question only asks for the type of roots, so we do not need to solve fully, the discriminant $b^2-4ac$ settles it. Identify the coefficients: $a=1$, $b=-6$ and $c=9$.

Squaring a negative $b$ still gives a positive value, which is a common place to slip.

When the discriminant equals $0$, the equation has two real and equal roots (a repeated root). We can see this directly, since $x^2-6x+9=(x-3)^2$, giving $x=3$ twice.

A positive discriminant would instead give two distinct roots, and a negative one would give no real roots.

The phrase 'minimum value' is the signal to complete the square, because the completed form shows the vertex at a glance. Halve the coefficient of $x$: half of $-6$ is $-3$, so we start from $(x-3)^2$.

Since $(x-3)^2 = x^2-6x+9$ carries an extra $+9$, we subtract it back so the value is unchanged:

The squared term $(x-3)^2$ is never negative; its smallest value is $0$, reached when the bracket is zero, that is when $x=3$.

So the minimum value of $f(x)$ is $2$, occurring at $x=3$. The turning point is therefore $(3,\,2)$, and it is a minimum because the coefficient of $x^2$ is positive, so the parabola opens upward.

Check by substitution: $f(3)=(3)^2-6(3)+11 = 9-18+11 = 2$, matching the value read from the completed square.

For a curve $y=ax^2+bx+c$, the axis of symmetry is the vertical line through the vertex, given directly by $x=-\dfrac{b}{2a}$. Here $a=1$, $b=-8$ and $c=10$, so read off the coefficients before substituting.

Simplify the double negative in the numerator, then divide.

When the two roots are known, the equation can be rebuilt directly from the factors $(x-\alpha)(x-\beta)=0$, keeping each root's own sign inside its bracket.

Expand the brackets and collect like terms.

Start by factorising to find the critical values, the $x$-values where the expression equals zero, since these are the points where the graph crosses the $x$-axis.

The coefficient of $x^2$ is positive, so the graph is an upward-opening parabola. It lies on or below the $x$-axis ($\le 0$) only on the interval between its two roots.

Two equal roots occur exactly when the discriminant $b^2-4ac$ is zero. Identify the coefficients $a=1$, $b=k$ and $c=16$, then set the discriminant to zero.

Solve this equation for $k$.