Problems : Quadratic Functions

Important Notes

Quadratic Function (Standard Form)	$f(x)=a x^{2}+b x+c, a \neq 0$
Graph	Parabola $a>0$ (opens upward) $a<0$ (opens downward)
Axis of Symmetry	$x=-\displaystyle\frac{b}{2 a}$
Vertex	$\left(-\displaystyle\frac{b}{2 a}, f\left(-\displaystyle\frac{b}{2 a}\right)\right)=\left(-\displaystyle\frac{b}{2 a},-\displaystyle\frac{b^{2}-4 a c}{4 a}\right)$
y-intercept	(0, c)
Discriminant	$b^{2}-4 a c$ $b^{2}-4 a c>0 \Rightarrow$ two $x$ intercepts (cuts $x-$ axis at two points) $b^{2}-4 a c=0 \Rightarrow$ one $x$ intercepts (touch $x$ -axis at one point $)$ $b^{2}-4 a c=0 \Rightarrow$ one $x$ intercepts (does not intersect $x$ -axis)
Quadratic Equation	$a x^{2}+b x+c=0, a \neq 0$
Quadratic Formula	$x=\displaystyle\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}$
Quadratic Function (Vertex Form)	$f(x)=a(x-h)^{2}+k, a \neq 0$
Vertex	$(h, k)$
Axis of Symmetry (Vertex Form)	$x=h$
Quadratic Function (Intercept Form when discriminant $>0$)	$f(x)=a(x-p)(x-q), a \neq 0$
$X$ -intercept points	$(p, 0)$ and $(q, 0)$
Axis of Symmetry	$x=\displaystyle\frac{p+q}{2}$
Quadratic Inequality	$a x^{2}+b x+c>0$ $a x^{2}+b x+c \geq 0$ $a x^{2}+b x+c<0$ $a x^{2}+b x+c \leq 0$
$a>0$ and $b^2-4ac<0$	The graph does not cut the $x$ -axis. $y<0 \Rightarrow$ solution set $=\varnothing$ $y=0 \Rightarrow$ solution set $=\varnothing$ $y>0 \Rightarrow$ solution set $=\mathbb{R}$
$a<0$ and $b^2-4ac<0$	The graph does not cut the $x$ -axis. $y<0 \Rightarrow$ solution set $=\mathbb{R}$ $y=0 \Rightarrow$ solution set $=\varnothing$ $y>0 \Rightarrow$ solution set $=\varnothing$
$a>0$ and $b^2-4ac=0$	The graph touches the $x$ -axis. $y<0 \Rightarrow$ solution set $=\varnothing$ $y=0 \Rightarrow$ solution set $=\left\{-\displaystyle\frac{b}{2 a}\right\}$ $y>0 \Rightarrow$ solution set $=\mathbb{R} \backslash\left\{-\displaystyle\frac{b}{2 a}\right\}$
$a<0$ and $b^2-4ac=0$	The graph touches the $x$ -axis. $y<0 \Rightarrow$ solution set $=\mathbb{R} \backslash\left\{-\displaystyle\frac{b}{2 a}\right\}$ $y=0 \Rightarrow$ solution set $=\left\{-\displaystyle\frac{b}{2 a}\right\}$ $y>0 \Rightarrow$ solution set $=\varnothing$
$a>0$ and $b^2-4ac>0$	The graph cuts the $x$ -axis at two points. $y<0 \Rightarrow$ solution set $=\{x \mid p<x<q\}$ $y=0 \Rightarrow$ solution set $=\{p, q\}$ $y>0 \Rightarrow$ solution set $=\{x \mid x<p$ or $x>q\}$
$a<0$ and $b^2-4ac>0$	The graph cuts the $x$ -axis at two points. $y<0 \Rightarrow$ solution set $=\{x \mid x<p$ or $x>q\}$ $y=0 \Rightarrow$ solution set $=\{p, q\}$ $y>0 \Rightarrow$ solution set $=\{x \mid p<x<q\}$

Example (1)

The equation $k x^{2}+5 k x+3=0$, where $k$ is a constant, has no real roots. Prove that $k$ satisfies the inequality $0 \leq k < \displaystyle\frac{12}{25}$.

Solution

If $k=0,3=0$ is impossible.

Therefore $k x^{2}+5 k x+3=0$ has no real root when $k=0 \quad---(1)$

$k x^{2}+5 k x+3=0$

Since the equation has no real root, discriminant $ < 0 $.

$\therefore(5 k)^{2}-4 k(3)< 0 $

$25 k^{2}-12 k<0$

$k(25 k-12)< 0$

Dividing both sides with 25 ,

$k\left(k-\displaystyle\frac{12}{25}\right)<0$

$0 < k <\displaystyle\frac{12}{25} \quad---(2)$

By equations (1) and (2), $0 \leq k < \displaystyle\frac{12}{25}$

Example (2)

Prove that $x^{2}+8 x+20 \geqslant 4$ for all values of $x$.

Solution

$x^{2}+8 x+20$

$=x^{2}+2 x(4)+4^{2}+4$

$=(x+4)^{2}+4$

Since $(x+4)^{2} \geq 0$ for all $x \in \mathbb{R}$,

$(x+4)^{2}+4 \geq 4$ for all $x \in \mathbb{R}$

$\therefore x^{2}+8 x+20 \geq 4$ for all $x \in \mathbb{R}$

Example (3)

Find the suitable domain of the function $f(x)=1+3 x-2 x^{2}$ for which the curve of $f(x)$ lies completely above the line $y=-1$.

Solution

$f(x)=1+3 x-2 x^{2}$

By the problem, $f(x)>-1$.

$\therefore 1+3 x-2 x^{2}>-1$

$\therefore 2+3 x-2 x^{2}>0$

$\therefore(1+x)(4-x)>0$

$\therefore \quad-1< x < 4$

$\therefore \operatorname{dom}(f)=\{x \mid-1<x<4\}$

Example (4)

The ratio of the lengths $a: b$ in this line is the same as the ratio of the lengths $b: c$.

Solution

By the diagram, $a=b+c$

By the problem, $\displaystyle\frac{a}{b}=\displaystyle\frac{b}{c}$

$\therefore \displaystyle\frac{b+c}{b}=\displaystyle\frac{b}{c}$

$\therefore b^{2}=b c+c^{2}$

$\therefore b^{2}-b c-c^{2}=0$

Dividing both sides with $c^{2}$,

$\left(\displaystyle\frac{b}{c}\right)^{2}-\displaystyle\frac{b}{c}-1=0$

$\therefore \displaystyle\frac{b}{c}=\displaystyle\frac{1 \pm \sqrt{1+4}}{2}$

$\therefore \displaystyle\frac{b}{c}=\displaystyle\frac{1 \pm \sqrt{5}}{2}$

Since $b, c>0, \displaystyle\frac{b}{c}=\displaystyle\frac{1+\sqrt{5}}{2}$

Example (5)

Show that the infinite square root $\sqrt{1+\sqrt{1+\sqrt{1+\sqrt{1+\sqrt{1+\ldots}}}}}=\displaystyle\frac{1+\sqrt{5}}{2} .$ Solution Let $x=\sqrt{1+\sqrt{1+\sqrt{1+\sqrt{1+\sqrt{1+\ldots}}}}}$ Squaring both sides, $x^{2}=1+\sqrt{1+\sqrt{1+\sqrt{1+\sqrt{1+\ldots}}}}$ $\therefore x^{2}=1+x$ $\therefore x^{2}-x-1=0$ $\therefore x=\displaystyle\frac{1 \pm \sqrt{5}}{2}$ Since $x>0, x =\displaystyle\frac{1+\sqrt{5}}{2}$ $\therefore \quad \sqrt{1+\sqrt{1+\sqrt{1+\sqrt{1+\sqrt{1+\ldots}}}}}=\displaystyle\frac{1+\sqrt{5}}{2} .$

Exercises

1.	Use completing the square to prove that $3 n^{2}-4 n+10$ is positive for all values of $n$.
2.	Use completing the square to prove that $-n^{2}-2 n-3$ is negative for all values of $n$.
3.	Find the values of $k$ for which $x^{2}+6 x+k=0$ has two real solutions.
4.	Find the value of $t$ for which $2 x^{2}-3 x+t=0$ has exactly one solution.
5.	Given that the function $f(x)=s x^{2}+8 x+s$ has equal roots, find the value of the positive constant $s$.
6.	Find the range of values of $k$ for which $3 x^{2}-4 x+k=0$ has no real solutions.
7.	The function $g(x)=x^{2}+3 p x+(14 p-3)$, where $p$ is an integer, has two equal roots.
	(a)	Find the value of $p$.
	(b)	For this value of $p$, solve the equation $x^{2}+3 p x+(14 p-3)=0$.
8.	$h(x)=2 x^{2}+(k+4) x+k$, where $k$ is a real constant.
	(a)	Find the discriminant of $h(x)$ in terms of $k$.
	(b)	Hence or otherwise, prove that $h(x)$ has two distinct real roots for all values of $k$.
9.	The equation $p x^{2}-5 x-6=0$, where $p$ is a constant, has two distinct real roots. Prove that $p$ satisfies the inequality $p>-\displaystyle\frac{25}{24} $.

Answer Keys

$\begin{array}{ll} 1. & \text{Hint:}\ 3\left(n-\displaystyle\frac{2}{3}\right)^{2}+\displaystyle\frac{26}{3}\\\\ 2. & \text{Hint:}\ -(n+1)^2-2\\\\ 3. & k<9 \\\\ 4. & t=\displaystyle\frac{9}{8}\\\\ 5. & s=4\\\\ 6. & \left\{x\ |\ k>\displaystyle\frac{4}{3}\right\}\\\\ 7. & \text{(a)}\ p=6, \text{(b)}\ x=-9\\\\ 8. & \begin{array}{ll} \text{(a)}& \text{discriminant}=(k+4)^2-8k \\ \text{(b)}& \text{Hint} : (k+4)^2-8k=k^2+16>0\ \text{for all values of}\ k \end{array}\\\\ 9. & \text{Hint: discriminant} >0 \end{array}$