Important Notes
| Quadratic Function (Standard Form) | $f(x)=a x^{2}+b x+c, a \neq 0$ |
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| 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 $x>0, x =\displaystyle\frac{1+\sqrt{5}}{2}$ $\therefore \quad \sqrt{1+\sqrt{1+\sqrt{1+\sqrt{1+\sqrt{1+\ldots}}}}}=\displaystyle\frac{1+\sqrt{5}}{2} .$ |
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Exercises
| 1. | Use completing the square to prove that $3 n^{2}-4 n+10$ is positive for all values of $n$. | |
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| 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
| $\begin{array}{ll} 1. & \text{Hint:}\ 3\left(n-\displaystyle\frac{2}{3}\right)^{2}+\displaystyle\frac{26}{3}\\\\ 2. & \text{Hint:}\ -(n+1)^2-2\\\\ 3. & k<9 \\\\ 4. & t=\displaystyle\frac{9}{8}\\\\ 5. & s=4\\\\ 6. & \left\{x\ |\ k>\displaystyle\frac{4}{3}\right\}\\\\ 7. & \text{(a)}\ p=6, \text{(b)}\ x=-9\\\\ 8. & \begin{array}{ll} \text{(a)}& \text{discriminant}=(k+4)^2-8k \\ \text{(b)}& \text{Hint} : (k+4)^2-8k=k^2+16>0\ \text{for all values of}\ k \end{array}\\\\ 9. & \text{Hint: discriminant} >0 \end{array}$ |
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