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

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

Introduction to Coordinate Geometry: Exercise (1.1) - Solution

1.           Draw a set of coordinate axes. Locate the points, $A(2,3)$, $B(2, -4)$ and $C(-4, 3)$. Label each point with its coordinates. Determine whether each of the line segments $AB, BC$ and $CA$ is horizontal or vertical.

Show/Hide Solution

• $AB$ is a vertical line.
• $BC$ is neither horizontal nor vertical line.
• $AC$ is a horizontal line.

2.           Find the missing coordinates in the following table if $M$ is the midpoint of points $P$ and $Q$. $$\begin{array}{|c|c|c|} \hline P & Q & M \\ \hline (2,6) & & (3,3) \\ \hline (3,2) & (-3,-1) & \\ \hline & (0,-1) & (-3,2) \\ \hline (1,5) & & (2.5,3.5) \\ \hline \end{array}$$

Show/Hide Solution

(i) Let the coordinates of the point $Q$ be $(a, b)$.

Since $M$ is the midpoint of $P$ and $Q$, using midpoint formula,

$(3,3) = \left(\displaystyle\frac{a+2}{2},\displaystyle \frac{b+6}{2}\right)$

$\therefore\ \displaystyle \frac{a+2}{2} = 3\ \text{and}\ \displaystyle \frac{b+6}{2} = 3$

$\therefore\ a=4\ \text{and}\ b = 0$

$\therefore\ Q=(4,0)$

(ii) Let the coordinates of the point $M$ be $(x, y)$.

Since $M$ is the midpoint of $P$ and $Q$, using midpoint formula,

$\begin{aligned} (x,y) &= \left(\frac{3-3}{2},\displaystyle \frac{2-1}{2}\right)\\\\ \therefore\ (x,y) &=\left(0,\frac{1}{2}\right) \end{aligned}$

(iii) Let the coordinates of the point $P$ be $(x, y)$.

Since $M$ is the midpoint of $P$ and $Q$, using midpoint formula,

$(-3,2) = \left(\displaystyle\frac{x+0}{2},\displaystyle \frac{y-1}{2}\right)$

$\therefore\ \displaystyle \frac{x}{2} = -3\ \text{and}\ \displaystyle \frac{y-1}{2} = 2$

$\therefore\ x=-6\ \text{and}\ y = 5$

$\therefore\ P=(-6,5)$

(iv) Let the coordinates of the point $Q$ be $(x, y)$.

Since $M$ is the midpoint of $P$ and $Q$, using midpoint formula,

$(2.5,3.5) = \left(\displaystyle\frac{x+1}{2},\displaystyle \frac{y+5}{2}\right)$

$\therefore\ \displaystyle \frac{x+1}{2} = 2.5\ \text{and}\ \displaystyle \frac{y+5}{2} = 3.5$

$\therefore\ x=4\ \text{and}\ y = 2$

$\therefore\ P=(4,2)$

3.           Find the coordinates of the midpoint and the length of the line segment joining these pairs of points. $$\begin{array}{l} \text{(a)}\ (0,0)\ \text{and}\ (4, -4) \\ \text{(b)}\ (1, 5)\ \text{and}\ (3,1) \\ \text{(c)}\ (-3, -3)\ \text{and}\ (0,0) \\ \text{(d)}\ (-1,3)\ \text{and}\ (5, 1) \\ \text{(e)}\ (-1,6)\ \text{and}\ (2, -2) \\ \text{(f)}\ (-3, -4)\ \text{and}\ (3, -1) \end{array}$$

Show/Hide Solution

$\begin{aligned} \text{(a)}\ & \text{The midpoint between } (0,0)\ \text{and }(4,-4)\\\\ &=\left( {\displaystyle \frac{{0+4}}{2},\displaystyle \frac{{0-4}}{2}} \right)\\\\ &=\left( {2,-2} \right)\\\\ & \text{length of segment}\\\\ &=\sqrt{(4-0)^2+(-4-0)^2}\\\\ &= 4\sqrt{2} \end{aligned}$

$\begin{aligned} \text{(b)}\ & \text{The midpoint between }(1,5)\ \text{and }(3,1)\\\\ &=\left( {\displaystyle \frac{{1+3}}{2},\displaystyle \frac{{5+1}}{2}} \right)\\\\ &=\left( {2,3} \right)\\\\ & \text{length of segment}\\\\ &=\sqrt{(3-1)^2+(1-5)^2}\\\\ &= 2\sqrt{5} \end{aligned}$

$\begin{aligned} \text{(c)}\ & \text{The midpoint between }(-3,-3)\ \text{and }(0,0)\\\\ &=\left( {\displaystyle \frac{{-3+0}}{2},\displaystyle \frac{{-3+0}}{2}} \right)\\\\ &=\left( {-\displaystyle \frac{{3}}{2},-\displaystyle \frac{{3}}{2}} \right)\\\\ & \text{length of segment}\\\\ &=\sqrt{(0+3)^2+(0+3)^2}\\\\ &=\sqrt{18}\\\\ &= 3\sqrt{2} \end{aligned}$

$\begin{aligned} \text{(d)}\ & \text{The midpoint between }(-1,3)\ \text{and }(5,1)\\\\ &=\left( {\displaystyle \frac{{-1+5}}{2},\displaystyle \frac{{3+1}}{2}} \right)\\\\ &=\left( {2,2} \right)\\\\ & \text{length of segment}\\\\ &=\sqrt{(5+1)^2+(1-3)^2}\\\\ &=\sqrt{40}\\\\ &= 2\sqrt{10} \end{aligned}$

$ \begin{aligned} \text{(e)}\ & \text{The midpoint between }(-1,6)\ \text{and }(2,-2)\\\\ &=\left( {\displaystyle \frac{{-1+2}}{2},\displaystyle \frac{{6-2}}{2}} \right)\\\\ &=\left(\displaystyle \displaystyle \frac{1}{2},2 \right)\\\\ &\text{length of segment}\\\\ &=\sqrt{(2+1)^2+(-2-6)^2}\\\\ &=\sqrt{73} \end{aligned}$

$ \begin{aligned} \text{(f)}\ &\text{The midpoint between }(-3,-4)\ \text{and }(3,-1)\\\\ &=\left( {\displaystyle \frac{{-3+3}}{2},\displaystyle \frac{{-4-1}}{2}} \right)\\\\ &=\left( {0,-\displaystyle \frac{{5}}{2}} \right)\\\\ & \text{length of segment}\\\\ &=\sqrt{(3+3)^2+(-1+4)^2}\\\\ &=\sqrt{45}\\\\ &= 3\sqrt{5} \end{aligned}$

4.           If $(1,0)$ is the midpoint of the line passing through the points $A(-5, 2)$ and $B(x, y)$, find the value of $x$ and of $y$.

Show/Hide Solution

Since $(1,0)$ is the midpoint of $A(-5,2)$ and $B(x,y)$, using midpoint formula,

$(1,0) = \left(\displaystyle \frac{-5+x}{2}, \displaystyle \frac{2+y}{2}\right)$

$\therefore\ \displaystyle \frac{-5+x}{2} = 1\ \text{and}\ \displaystyle \frac{2+y}{2} = 0$

$\therefore\ x=7\ \text{and}\ y = -2$

5.           Calculate the perimeter of given polygons correct to one decimal place.

(a) A triangle with vertices $P(-2, 3), Q(5, -4)$ and $R(1, 8)$.

(b) A parallelogram with vertices $A(-10, 1), B(6, -2), C(14, 4)$ and $D(-2, 7)$.

(c) A trapezium with vertices $E(-6, -2), F(1, -2), G(0, 4)$ and $H(-5, 4)$.

Show/Hide Solution

(a) $ P=(-2,3),Q=(5,-4),R=(1,8)$

Since the distance between $(x_1, y_1)$ and $(x_2, y_2)$ is $\sqrt{\left( x_2-x_1 \right)^2 + \left(y_2-y_1 \right)^2},$

$\begin{aligned} PQ&=\sqrt{(5+2)^2+(-4-3)^2}\\\\ &=\sqrt{98}\\\\ &=9.9\\\\ QR &=\sqrt{(1-5)^2+(8+4)^2}\\\\ &=\sqrt{160}\\\\ &=12.7\\\\ PR&=\sqrt{(1+2)^2+(8-3)^2}\\\\ &=\sqrt{34}\\\\ &=5.8 \end{aligned}$

$\therefore\ PQ+QR+PR=28.4$

$\therefore$ the perimeter of $\triangle PQR = 28.4$ units.

(b) $A=(-10, 1), B=(6, -2), C=(14, 4), D=(-2, 7)$

Since the distance between $(x_1, y_1)$ and $(x_2, y_2)$ is $\sqrt{\left( x_2-x_1 \right)^2 + \left(y_2-y_1 \right)^2},$

$\begin{aligned} AB &=\sqrt{(6+10)^2+(-2-1)^2}\\\\ &=\sqrt{265}\\\\ &=16.3\\\\ BC &=\sqrt{(14-6)^2+(4+2)^2}\\\\ &=\sqrt{100}\\\\ &=10.0 \end{aligned}$

Since $ABCD$ is a parallelogram, $CD=AB$ and $AD = BC$

$\therefore\ AB + BC + CD + AD =2(16.3) + 2(10)=52.6$

$\therefore$ the perimeter of triangle parallelogram $ABCD = 52.6$ units.

(c) $E=(-6, -2), F=(1, -2), G=(0, 4), H=(-5, 4)$

Since the distance between $(x_1, y_1)$ and $(x_2, y_2)$ is $\sqrt{\left( x_2-x_1 \right)^2 + \left(y_2-y_1 \right)^2},$

$\begin{aligned} EF &=\sqrt{(1+6)^2+(-2+2)^2}\\\\ &=\sqrt{7^2}\\\\ &=7.0\\\\ FG &=\sqrt{(0-1)^2+(4+2)^2}\\\\ &=\sqrt{37}\\\\ &=6.1\\\\ GH &=\sqrt{(-5-0)^2+(4-4)^2}\\\\ &=\sqrt{5^2}\\\\ &=5.00\\\\ EH &=\sqrt{(-5+6)^2+(4+2)^2}\\\\ &=\sqrt{37}\\\\ &=6.1\\\\ \end{aligned}$

$\therefore\ AB + BC + CD + AD =7.0+6.1+5.0+6.1=24.2$

6.           A circle has centre $(2, 1$). Find the coordinates of the endpoint of a diameter if one endpoint is $(7, 1)$.

Show/Hide Solution

Let the other endpoint be $(x, y)$.

Hence, $(2, 1)$ is the midpoint between $(x, y)$ and $(7,1)$

Using midpoint formula,

$(2, 1) = \left(\frac{x+7}{2}, \frac{y+1}{2}\right)$

$\therefore\ \frac{x+7}{2} = 2\ \text{and}\ \frac{y+1}{2} = 1$

$\therefore\ x = -3\ \text{and}\ y = 1$

Hence the other endpoint is $(-3, 1).$

7.           $\triangle KLM$ has vertices $K(-5,18)$, $L(10,14)$ and $M(-5, -10)$.

(a) Find the length of each side.

(b) Find the perimeter of $\triangle KLM$.

(c) Find the area of $\triangle KLM$.

Show/Hide Solution

Since the distance between $(x_1, y_1)$ and $(x_2, y_2)$ is $\sqrt{\left( x_2-x_1 \right)^2 + \left(y_2-y_1 \right)^2},$

$\begin{aligned} KL &=\sqrt{(10+5)^2+(14-18)^2}\\\\ &=\sqrt{241}\\\\ &=15.5\\\\ LM &=\sqrt{(-5-10)^2+(-10-14)^2}\\\\ &=\sqrt{801}\\\\ &=3\sqrt{89}\\\\ &=28.3\\\\ KM &=\sqrt{(-5+5)^2+(-10-18)^2}\\\\ &=\sqrt{28^2}\\\\ &=28\\\\ \text{The}\ & \text{perimeter of}\ \triangle KLM \\\\ & = KL+LM+KM\\\\ & = 15.5+28.3+28\\\\ &= 71.8\ \text{units} \end{aligned}$

Since $K$ and $M$ have the same $x$-coordinate, $KM$ is a vertical line.

Draw $LN\bot KM$.

Hence the coordinates of $N$ is $(-5,14)$.

$\begin{aligned} \therefore\ LN &=\sqrt{(-5-10)^2+(14-14)^2}\\\\ &=\sqrt{15^2}\\\\ &=15\\\\ \text{The}\ & \text{area of}\ \triangle KLM \\\\ & = \frac{1}{2}\cdot KM\cdot LN\\\\ & = \frac{1}{2}\cdot 28\cdot 15\\\\ & = 210\ \text{sq-units} \end{aligned}$

Exercise (5.5) : Solving Linear Quadratic Systems Algebraically

1.           Find the solution set of each of the systems of equations:

              (a)   $x^2-y^2=9$

                     $x+y=1$

Show/Hide Solution

$\begin{array}{l}{{x}^{2}}-{{y}^{2}}=9\cdot \cdot \cdot \cdot \cdot \cdot \cdot (1)\\\\x+y=1\ \ \ \cdot \cdot \cdot \cdot \cdot \cdot \cdot (2)\\\\\text{From equation }(2)\text{,we have}\\\\y=1-x\ \ \ \cdot \cdot \cdot \cdot \cdot \cdot \cdot (3)\\\\\text{Substituting }y=1-x\text{ in equation }(1)\text{,}\\\\{{x}^{2}}-{{\left( {1-x} \right)}^{2}}=9\\\\{{x}^{2}}-\left( {1-2x+{{x}^{2}}} \right)=9\\\\{{x}^{2}}-1+2x-{{x}^{2}}=9\\\\2x=10\\\\x=5\\\\\text{Substituting }x=5\text{ in equation }(3),\\\\y=1-5=-4\\\\\therefore \ \ \text{Solution set}=\left\{ {\left( {5,-4} \right)} \right\}\text{ }\end{array}$

              (b)   $y=\displaystyle \frac{8}{x}$

                     $y=7+x$

Show/Hide Solution

$\begin{array}{l}y=\displaystyle \frac{8}{x}\cdot \cdot \cdot \cdot \cdot \cdot \cdot (1)\\\\y=7+x\ \ \ \cdot \cdot \cdot \cdot \cdot \cdot \cdot (2)\\\\\text{From equation }(2)\text{,we have}\\\\\displaystyle \frac{8}{x}=7+x\\\\\therefore \ {{x}^{2}}+7x=8\\\\{{x}^{2}}+7x-8=0\\\\(x+8)(x-1)=0\\\\x=-8\ \text{or}\ x=1\\\\\text{When}\ x=-8,\ y=7-8=-1\\\\\text{When}\ x=1,\ y=7+1=8\\\\\therefore \ \ \text{Solution set}=\left\{ {\left( {-8,-1} \right),\left( {1,8} \right)} \right\}\text{ }\end{array}$

              (c)   $x^2+5x+y=4$

                     $x+y=8$

Show/Hide Solution

$\begin{array}{l}{{x}^{2}}+5x+y=4\cdot \cdot \cdot \cdot \cdot \cdot \cdot (1)\\\\x+y=8\ \ \ \ \ \ \ \cdot \cdot \cdot \cdot \cdot \cdot \cdot (2)\\\\\text{From equation }(2)\text{,we have}\\\\y=8-x\ \ \ \ \ \ \cdot \cdot \cdot \cdot \cdot \cdot \cdot (3)\\\\\text{Substituting}\ y=8-x\text{ in equation }(1),\\\\{{x}^{2}}+5x+8-x=4\\\\{{x}^{2}}+4x+4=0\\\\{{(x+2)}^{2}}=0\\\\x=-2\\\\\text{Substituting}\ x=-2\text{ in equation }(3),\\\\y=8-(-2)=10\\\\\therefore \ \ \text{Solution set}=\left\{ {\left( {-2,10} \right)} \right\}\text{ }\end{array}$

2.           The sum of squares of two numbers is 58. If the first number and twice the second add up to 13, find the numbers.

Show/Hide Solution

$\begin{array}{l}\text{Let}\ \text{the two numbers be }x\ \text{and }y.\\\\\text{By the problem,}\\\\{{x}^{2}}+{{y}^{2}}=58\cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \ (1)\\\\x+2y=13\ \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot (2)\\\\\text{From equation }(2)\text{,we have}\\\\x=13-2y\ \ \ \ \ \ \cdot \cdot \cdot \cdot \cdot \cdot \cdot (3)\\\\\text{Substituting}\ x=13-2y\text{ in equation }(1),\\\\{{\left( {13-2y} \right)}^{2}}+{{y}^{2}}=58\\\\169-52y+4{{y}^{2}}+{{y}^{2}}=58\\\\169-52y+5{{y}^{2}}=58\\\\\therefore \ \ 5{{y}^{2}}-52y+111=0\\\\(5y-37)(y-3)=0\\\\y=\displaystyle \frac{{37}}{5}\ \text{or}\ y=3\\\\\text{When}\ y=\displaystyle \frac{{37}}{5},x=13-2\left( {\displaystyle \frac{{37}}{5}} \right)\text{=}-\displaystyle \frac{9}{5}\text{ }\\\\\text{When}\ y=3,x=13-2\left( 3 \right)\text{=}7\text{ }\\\\\therefore \ \ \text{The two numbers are }-\displaystyle \frac{9}{5}\ \text{and }\displaystyle \frac{{37}}{5}\ \text{or 7 and 3}\text{.}\end{array}$

3.           The sum of the reciprocals of two positive numbers is $\displaystyle \frac{7}{36}$ and the product of the numbers is 108. Find the numbers.

Show/Hide Solution

$\begin{array}{l}\text{Let}\ \text{the two positive numbers be }x\ \text{and }y.\\\\\text{By the problem,}\\\\\displaystyle \frac{1}{x}+\displaystyle \frac{1}{y}=\displaystyle \frac{7}{{36}}\cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \ (1)\\\\xy=108\ \ \ \ \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot (2)\\\\\text{From equation }(2)\text{,we have}\\\\y=\displaystyle \frac{{108}}{x}\ \ \ \ \ \ \cdot \cdot \cdot \cdot \cdot \cdot \cdot (3)\\\\\text{Substituting}\ y=\displaystyle \frac{{108}}{x}\text{ in equation }(1),\\\\\displaystyle \frac{1}{x}+\displaystyle \frac{1}{{\displaystyle \frac{{108}}{x}}}=\displaystyle \frac{7}{{36}}\\\\\displaystyle \frac{1}{x}+\displaystyle \frac{x}{{108}}=\displaystyle \frac{7}{{36}}\\\\\text{Multiplying both sides of equation by}\ 108x,\\\\108+{{x}^{2}}=21x\\\\\therefore \ \ {{x}^{2}}+21x-108=0\\\\(x-9)(x-12)=0\\\\x=9\ \text{or}\ x=12\\\\\text{When}\ x=9,y=\displaystyle \frac{{108}}{9}=12\\\\\text{When}\ x=12,y=\displaystyle \frac{{108}}{{12}}=9\\\\\therefore \ \ \text{The two positive numbers are }9\text{ and 12}\text{.}\end{array}$