Quadratic Functions, Worked Examples (medium)

Two distinct real roots occur when the discriminant is positive: $b^2-4ac > 0$. Here $a=1$, $b=p$ and $c=4$.

$$p^2 - 4(1)(4) > 0 \;\Rightarrow\; p^2 - 16 > 0$$

Factorise the left side as a difference of two squares:

$$(p-4)(p+4) > 0$$

This product is positive outside the roots $p=-4$ and $p=4$. So the solution is:

$$p < -4 \quad \text{or} \quad p > 4$$

Check $p=5$: discriminant $=25-16=9>0$, two distinct roots, consistent. Check $p=0$: discriminant $=-16<0$, no real roots, correctly excluded.

For $ax^2+bx+c=0$, the sum of roots is $-\frac{b}{a}$ and the product is $\frac{c}{a}$. Here $a=1$, $b=-5$, $c=2$, so:

$$\alpha+\beta = -\frac{-5}{1} = 5, \qquad \alpha\beta = \frac{2}{1} = 2$$

Now find the sum of the new roots $\alpha+1$ and $\beta+1$:

$$(\alpha+1)+(\beta+1) = \alpha+\beta+2 = 5+2 = 7$$

And the product of the new roots:

$$(\alpha+1)(\beta+1) = \alpha\beta + \alpha + \beta + 1 = 2 + 5 + 1 = 8$$

A quadratic equation with these roots is $x^2 - (\text{sum})x + (\text{product}) = 0$:

$$x^2 - 7x + 8 = 0$$

The new sum $7$ and product $8$ are both consistent with shifting each original root up by $1$.

First find where the expression equals zero by factorising:

$$x^2 - x - 6 = (x-3)(x+2)$$

So the critical values are $x=3$ and $x=-2$. The graph of $y=x^2-x-6$ is a parabola opening upward, so $y$ is negative between the two roots.

$$-2 < x < 3$$

Check inside with $x=0$: $0-0-6=-6<0$, satisfied. Check outside with $x=4$: $16-4-6=6>0$, not satisfied, confirming the range.

Complete the square by halving the coefficient of $x$, which is $-6$, to get $-3$, then adjust the constant so the expression stays equal.

$$f(x) = x^2 - 6x + 11 = (x-3)^2 - 9 + 11$$

Simplify the constant terms to reach the completed-square form.

$$f(x) = (x-3)^2 + 2$$

A tangent touches the curve at exactly one point, so substitute the line into the curve's equation and set the resulting discriminant to zero.

$$x^2 + 4x + 7 = 2x + k \;\Rightarrow\; x^2 + 2x + (7-k) = 0$$

For one repeated root, $b^2-4ac=0$ with $a=1$, $b=2$, $c=7-k$.

$$(2)^2 - 4(1)(7-k) = 0 \;\Rightarrow\; 4 - 28 + 4k = 0 \;\Rightarrow\; k = 6$$

This equation is quadratic in $x^2$: let $y=x^2$, then factorise and solve for $y$.

$$y^2 - 13y + 36 = (y-4)(y-9) = 0 \;\Rightarrow\; y = 4 \text{ or } y = 9$$

Substitute back $x^2=y$ and take square roots of each value.

$$x^2 = 4 \;\Rightarrow\; x=\pm2, \qquad x^2=9 \;\Rightarrow\; x=\pm3$$

Since the perimeter is $24$ m, the length is $12-x$, so the area is:

$$A = x(12-x) = 12x - x^2$$

Complete the square to find the maximum, since the coefficient of $x^2$ is negative.

$$A = -(x^2-12x) = -[(x-6)^2 - 36] = 36 - (x-6)^2$$

A quadratic with minimum point $(2,-9)$ can be written in vertex form $f(x) = a(x-2)^2 - 9$. Substitute the point $(0,-1)$ to find $a$.

$$a(0-2)^2 - 9 = -1 \;\Rightarrow\; 4a = 8 \;\Rightarrow\; a = 2$$

Substitute $a=2$ back and expand to the general form.

$$f(x) = 2(x-2)^2 - 9 = 2x^2 - 8x - 1$$