Quiz

Hint

We know that a nonlinear function’s graph is not a straight line. So among the given graphs, (a), (b), and (c) are nonlinear whereas (d) is a linear function.

Hint

We can start by rearranging each equation so that \(y\) is on one side:
\(
\begin{aligned}
& -x^2+y+3 x=7 \rightarrow y=x^2-3 x+7 \\
& 5=-y-x^2+7 x \rightarrow y=-x^2+7 x-5
\end{aligned}
\)
Now that both equations are equal to \(\mathrm{y}\), we can set the right sides equal to each other and solve for \(\mathrm{x}\) :
\(
\begin{aligned}
& x^2-3 x+7=-x^2+7 x-5 \rightarrow 2 x^2-10 x+12=0 \\
& 2\left(x^2-5 x+6\right)=0 \rightarrow 2(x-2)(x-3)=0 \\
& x=2, x=3
\end{aligned}
\)
Our last step is to plug these values of \(x\) into either equation to find the \(y\) values of our solutions:
\(
\begin{aligned}
& y=x^2-3 x+7=(2)^2-3(2)+7=5 \\
& y=x^2-3 x+7=(3)^2-3(3)+7=7
\end{aligned}
\)
So our solutions are the following two points:
\(
(2,5),(3,7)
\)

Hint

Our first step is to solve the bottom equation for \(y\)
\(
y=x+3
\)
We can now substitute it into the top equation:
\(
x^2+(x+3)^2=117
\)
and solve for our \(x\) values:
\(
\begin{aligned}
& x^2+x^2+6 x+9=117 \\
& 2 x^2+6 x+9-117=0 \\
& 2 x^2+6 x-108=0 \\
& 2\left(x^2+3 x-54\right)=0 \\
& 2(x+9)(x-6)=0
\end{aligned}
\)
our values are then: \(x=6,-9\)
We can now plug our values into the bottom equation we had solved for \(y\) and arrive at our solutions:
\(
y=6+3=9 \quad y=-9+3=-6
\)
So, our solutions are \((6,9),(-9,-6)\)

Hint

Our first step is to solve the bottom equation for \(x^2\) :
\(
\begin{aligned}
& x^2(1+3 y)=x^4 \\
& (1+3 y)=\frac{x^4}{x^2} \\
& (1+3 y)=x^2
\end{aligned}
\)
so we can substitute it into the top equation:
\(
\begin{aligned}
& (1+3 y)+y^2=5 \\
& 1+3 y+y^2-5=0 \\
& y^2+3 y-4=0 \\
& (y-1)(y+4)=0 \\
& y=1,-4
\end{aligned}
\)
Now we can plug in our \(y\)-values into the bottom equation to find our \(\mathrm{x}\)-values:
\(
\begin{aligned}
& (1+3(1))=x^2 \\
& 4=x^2 \\
& x=\pm 2
\end{aligned}
\)
\(
\begin{aligned}
& (1+3(-4))=x^2 \\
& 1-12=x^2 \\
& -11=x^2
\end{aligned}
\)
Remember we cannot take a square root of a negative number without getting an imaginary number. As such, we’ll just focus on the \(x=\pm 2\) values.
Our solution is then: \((2,1),(-2,1)\)

Hint

The first step is to solve the bottom equation for \(y\) :
\(y=\sqrt{36}=\pm 6\) since our question specifies for \(y>0\) we just focus on \(y=6\)
We now substitute this equation into the top equation:
\(
\begin{aligned}
& 4(6)=x^2+2 x \\
& 0=x^2+2 x-24 \\
& 0=(x+6)(x-4) \\
& x=-6,4
\end{aligned}
\)
we can now plug in our \(x\)-values into the bottom equation to find our \(y\)-values:
\(
\begin{array}{lc}
4 y=(-6)^2+2(-6) & 4 y=(4)^2+2(4) \\
4 y=36-12 & 4 y=16+8 \\
4 y=24 & 4 y=24 \\
y=6 & y=6
\end{array}
\)
The solutions are then: \((-6,6),(4,6)\)

Hint

We can solve this equation by using substitution since the bottom equation is already solved for \(y^2\). Substituting the bottom equation into the top we get:
\(
\left(x^2-2\right)+x^2+6 x=18
\)
We then solve the equation for our \(x\) values:
\(
\begin{aligned}
& \left(x^2-2\right)+x^2+6 x-18=0 \\
& 2 x^2+6 x-20=0 \\
& 2\left(x^2+3 x-10\right)=0 \\
& 2(x-2)(x+5)=0 \\
& x=2,-5
\end{aligned}
\)
Finally, we substitute our values into the bottom equation to get our \(y\) values:
\(
\begin{array}{ll}
y^2=(2)^2-2 & y^2=(-5)^2-2 \\
y^2=2 & y^2=23 \\
y=\pm \sqrt{2} & y=\pm \sqrt{23}
\end{array}
\)
Our different solutions are then: \((2, \sqrt{2}),(2,-\sqrt{2}),(-5, \sqrt{23}),(-5,-\sqrt{23})\)

Hint

\(y=x^2\) is a parabola
\(x y=4\) is a hyperbola.
\(
\text { We see that the intersection point (the graphical solution) is at approximately }(1.5,2.5) \text {. }
\)

Hint

We see that the solutions are approximately
(0.5,1.8)(0.5,1.8) & (4.2,−0.9)(4.2,−0.9).

Hint

By the distance formula, we have
i.e.
\(
\begin{aligned}
& \sqrt{(x-h)^2+(y-k)^2}=r \\
& (x-h)^2+(y-k)^2=r^2
\end{aligned}
\)

\(
\text { Here } h=k=0 \text {. Therefore, the equation of the circle is } x^2+y^2=r^2 \text {. }
\)

Hint

Here \(h=-3, k=2\) and \(r=4\). Therefore, the equation of the required circle is
\(
(x+3)^2+(y-2)^2=16
\)

Hint

The given equation is
\(
\left(x^2+8 x\right)+\left(y^2+10 y\right)=8
\)
Now, completing the squares within the parenthesis, we get
\(
\begin{aligned}
& \left(x^2+8 x+16\right)+\left(y^2+10 y+25\right)=8+16+25 \\
& (x+4)^2+(y+5)^2=49 \\
& \{x-(-4)\}^2+\{y-(-5)\}^2=7^2
\end{aligned}
\)
Therefore, the given circle has centre at \((-4,-5)\) and radius 7.

Hint

Let \(\mathrm{F}\) be the focus and \(l\) the directrix. Let FM be perpendicular to the directrix and bisect FM at the point O. Produce MO to X. By the definition of parabola, the mid-point \(\mathrm{O}\) is on the parabola and is called the vertex of the parabola. Take \(O\) as origin, OX the \(x\)-axis and OY perpendicular to it as the \(y\)-axis. Let the distance from the directrix to the focus be \(2 a\). Then, the coordinates of the focus are \((a, 0)\), and the equation of the directrix is \(x+a=0\) as in above. Let \(\mathrm{P}(x, y)\) be any point on the parabola such that \(\mathrm{PF}=\mathrm{PB} \dots(1)\)
where \(\mathrm{PB}\) is perpendicular to \(l\). The coordinates of \(\mathrm{B}\) are \((-a, y)\). By the distance formula, we have
\(
\mathrm{PF}=\sqrt{(x-a)^2+y^2} \text { and } \mathrm{PB}=\sqrt{(x+a)^2}
\)
Since \(\mathrm{PF}=\mathrm{PB}\), we have
\(
\sqrt{(x-a)^2+y^2}=\sqrt{(x+a)^2}
\)
i.e. \((x-a)^2+y^2=(x+a)^2\)
or \(x^2-2 a x+a^2+y^2=x^2+2 a x+a^2\)
or \(y^2=4 a x(a>0)\).

Hence, any point on the parabola satisfies
\(
y^2=4 a x \dots(2)
\)
Conversely, let \(\mathrm{P}(x, y)\) satisfy the equation (2)
\(
\begin{aligned}
\mathrm{PF} & =\sqrt{(x-a)^2+y^2}=\sqrt{(x-a)^2+4 a x} \\
& =\sqrt{(x+a)^2}=\mathrm{PB} \dots(3)
\end{aligned}
\)
and so \(\mathrm{P}(x, y)\) lies on the parabola.
Thus, from (2) and (3) we have proved that the equation to the parabola with vertex at the origin, focus at \((a, 0)\) and directrix \(x=-a\) is \(y^2=4 a x\).

Hint

Since the vertex is at \((0,0)\) and the focus is at \((0,2)\) which lies on \(y\)-axis, the \(y\)-axis is the axis of the parabola. Therefore, equation of the parabola is of the form \(x^2=4 a y\). thus, we have
\(
x^2=4(2) y \text {, i.e., } x^2=8 y
\)

Hint

Since the foci is on y-axis, the equation of the hyperbola is of the form \(\frac{y^2}{a^2}-\frac{x^2}{b^2}=1\)
Since vertices are \(\left(0, \pm \frac{\sqrt{11}}{2}\right), a=\frac{\sqrt{11}}{2}\)
Also, since foci are \((0, \pm 3) ; c=3\) and \(b^2=c^2-a^2=\frac{25}{4}\).
Therefore, the equation of the hyperbola is
\(
\frac{y^2}{\left(\frac{11}{4}\right)}-\frac{x^2}{\left(\frac{25}{4}\right)}=1, \text { i.e., } 100 y^2-44 x^2=275
\)

Hint

The power of \(x\) in \(3 x^2\) and \(2 x\) are 2 and 1 respectively. So its highest power is 2 .
So it is a Nonlinear function

Hint

Step 1 The equations are already solved for \(y\).
Step 2 Substitute \(-2 x+3\) for \(y\) in Equation 1 and solve for \(x\).
\(
\begin{array}{rlrl}
-2 x+3 & =x^2+x-1 \\
3 & =x^2+3 x-1 \\
0 & =x^2+3 x-4 \\
0 & =(x+4)(x-1) \\
x+4 & =0 \quad \text { or } & x-1=0 \\
x & =-4 \quad \text { or } \quad x=1
\end{array}
\)
Step 3 Substitute \(-4\) and 1 for \(x\) in Equation 2 and solve for \(y\).
\(
\begin{array}{rlrlrl}
y & =-2(-4)+3 & \text { Substitute for } x \text { in Equation 2. } & & y & =-2(1)+3 \\
& =11 & \text { Simplify. } & & =1
\end{array}
\)
So, the solutions are \((-4,11)\) and \((1,1)\).

Hint

Subtract Equation 2 from Equation 1 .
\(
\begin{aligned}
0=x^2 \quad+6
\end{aligned}
\)

Solve for \(x\).
\(
\begin{aligned}
0 & =x^2+6 \\
-6 & =x^2
\end{aligned}
\)
The square of a real number cannot be negative. So, the system has no real solutions.

Hint

\(
\text { Let’s set up a system of non-linear equations: }\left\{\begin{array}{l}
x-y=3 \\
x^3+y^3=407
\end{array}\right.
\)
\(
\text { Substituting the } y \text { from the first equation into the second and solving yields: }
\)
\(
\begin{aligned}
x^3+(x-3)^3 & =407 \\
x^3+(x-3)\left(x^2-6 x+9\right) & =407 \\
x^3+x^3-6 x^2+9 x-3 x^2+18 x-27 & =407 \\
2 x^3-9 x^2+27 x-434 & =0
\end{aligned}
\)
\(
x=7 \text { works, and to find } y \text {, we use } y=x-3 \text {. When } x=7, y=4 \text {. The two numbers are } 4 \text { and } 7.
\)

Hint

Quadratic Formula to solve:
\(
\begin{aligned}
& \left\{\begin{array}{l}
d(t)=150+75 t-1.2 t^2 \\
d(t)=4 t^2
\end{array}\right. \\
& 150+75 t-1.2 t^2=4 t^2 \\
& 5.2 t^2-75 t-150=0 \\
& t=\frac{-(-75) \pm \sqrt{(-75)^2-4(5.2)(-150)}}{2(5.2)} \\
& t \approx 16.20 \\
&
\end{aligned}
\)
Note that we only want the positive value for \(t\), so in \(16.2\) seconds, the police car will catch up with Lacy.

Hint

Let the speed of the stream be \(x \mathrm{kmph}\).
We know that when the boat travels upstream, the relative speed between the boat and the stream \(=(18-x) \mathrm{kmph}\)
When the boat travels downstream, the relative speed between the boat and the stream \(=(18+x) \mathrm{kmph}\)
The time required to go upstream \(=\frac{\text { Distance }}{\text { Speed }}=\frac{24}{18-x}\) hours
Similarly, the time required to go downstream \(=\frac{\text { Distance }}{\text { Speed }}=\frac{24}{18+x}\) hours
Now, according to the question, \(\frac{24}{18-x}-\frac{24}{18+x}=1 \Rightarrow 24\left[\frac{1}{18-x}-\frac{1}{18+x}\right]=1 \Rightarrow\left[\frac{1}{18-x}-\frac{1}{18+x}\right]=\frac{1}{24}\)
To solve the for \(x\) we need to convert the above equation into the standard form of the quadratic equation.
\(
\begin{aligned}
& \Rightarrow\left[\frac{(18+x)-(18-x)}{(18-x)(18+x)}\right]=\frac{1}{24} \Rightarrow\left[\frac{18+x-18+x}{(18-x)(18+x)}\right]=\frac{1}{24} \\
& \Rightarrow\left[\frac{2 x}{18^2-x^2}\right]=\frac{1}{24} \text { (using the algebraic identity } a^2-b^2=(a+b)(a-b)
\end{aligned}
\)
Now, cross multiplying we have, \(\Rightarrow 48 x=324-x^2\)
\(
\Rightarrow x^2+48 x-324=0 \Rightarrow x^2+2 \times 24 \times x-324=0
\)
Now, to make the perfect square, we need to add and subtract \((24)^2\) from LHS
\(
\Rightarrow x^2+2 \times 24 \times x+(24)^2-(24)^2-324=0 \Rightarrow(x+24)^2-(24)^2-324=0
\)
Transferring \(-(24)^2-324\) from LHS to RHS we have,
\(
\Rightarrow(x+24)^2=+324+(24)^2
\)
Taking square root on both sides we have,
\(
\begin{aligned}
& \Rightarrow x+24=\pm \sqrt{900} \\
& \Rightarrow x=\pm 30-24 \\
& \Rightarrow x=+30-24 =6\\
& \Rightarrow x=-30-24=-54
\end{aligned}
\)
We will consider the positive value of \(x\) as speed can not be negative.
Hence, the speed of the stream is \(6 \mathrm{kmph}\).

Hint

Let us assume breadth is \(x\), and the length is \((x+25)\). According to the question, \(x(x+25)=300 \Rightarrow x^2+25 x-300=0\)
\(\Rightarrow\) From the given quadratic equation \(a=1, b=25, c=-300\)
Quadratic equation formula is given by \(x=\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}\)
\(x=\frac{-(25) \pm \sqrt{(25)^2-4 \times 1 \times(-300)}}{2 \times 1}=\frac{-25 \pm \sqrt{1825}}{2}\)
\(x=\frac{-25+\sqrt{1825}}{2}=8.86\) (approx) or \(x=\frac{-25-\sqrt{1825}}{2}=-33.86\) (approx) We will avoid the negative value of \(x\) as breadth can not be negative.
Hence, the breadth is, \(x=8.86\) (approx) and the length is \(x+25=8.86\) (approx) \(+25=33.86\) (approx)

Hint

Let the two odd integers be \(x, x+2\). We have:
Now, according to the statement,
\(
\begin{aligned}
& x(x+2)=63 \Rightarrow x^2+2 x-63=0 \\
& \Rightarrow x^2+(9-7) x-63=0 \Rightarrow x^2+9 x-7 x-63=0 \\
& \Rightarrow x(x+9)-7(x+9)=0 \Rightarrow(x+9)(x-7)=0
\end{aligned}
\)
So, \(x=-9\) or, \(x=7\)
Here, we need to find the positive odd integers. So, avoid the negative value of \(x\).
Hence, the two odd consecutive integers are \(x=7\) and \(x+2=7+2=9\).

Hint

Let her present age be \(x\).
Two years ago, her age was \((x-2)\).
Her age after four years from now is \((x+4)\).
One more than twice her present age is \((1+2 x)\).
According to the statement,
\(
\begin{aligned}
& (x-2)(x+4)=1+2 x x^2+4 x-2 x-8=1+2 x \\
& x^2+2 x-2 x-8-1=0 \\
& x^2-9=0 x^2=9
\end{aligned}
\)
Thus, \(x=\pm 3\)
Now, age can not be negative.
Hence, Rime’s present age is 3 years

Hint

Let the initial speed of the car is \(x \mathrm{kmph}\).
Speed of the car after increasing the speed is \((x+10), \mathrm{kmph}\).
Distance is \(72 \mathrm{~km}\) (given).
Time taken to cover the distance is \(\left(\frac{72}{x}\right)\) hours
Time taken after increasing the speed is \(\left(\frac{72}{x+10}\right)\) hours
According to the question,
\(
\begin{aligned}
& \Rightarrow\left(\frac{1}{x}\right)-\left(\frac{1}{x+10}\right)=\frac{36}{60 \times 72} \\
& \Rightarrow\left[\frac{x+10-x}{x(x+10)}\right]=\frac{1}{120} \Rightarrow x^2+10 x-1200=0 \\
& \Rightarrow x^2+40 x-30 x-1200=0 \\
& \Rightarrow x(x+40)-30(x+40)=0 \\
& \Rightarrow(x+40)(x-30)=0 \Rightarrow x=-40 \text { or } x=30
\end{aligned}
\)