Solving word problems by forming equations

Let the smaller number be $x$ ($x$ an even natural number). We have the equation $x(x+2)=224\Leftrightarrow x^2+2x-224=0$. We have $\Delta'=1+224=225$, so $x=-1+15=14$ (satisfies the condition) or $x=-1-15=-16$ (rejected).

So the two numbers are $14$ and $16$.

Answer. $14$ and $16$.

Let the shortest side be $x$ (cm), $x>0$. By Pythagoras' theorem we have

$$x^2+(x+7)^2=169\Leftrightarrow2x^2+14x-120=0\Leftrightarrow x^2+7x-60=0.$$

We have $\Delta=49+240=289$, so $x=\dfrac{-7+17}{2}=5$ (satisfies the condition) or $x=\dfrac{-7-17}{2}=-12$ (rejected).

So the two shorter sides are $5$ cm and $12$ cm.

Answer. $5$ cm and $12$ cm.

Let the planned number of products per day be $x$ ($x\in\mathbb{N}^*$). From the problem we have the equation

$$\frac{600}{x}-\frac{600}{x+10}=2\Leftrightarrow6000=2x(x+10)\Leftrightarrow x^2+10x-3000=0.$$

We have $\Delta'=25+3000=3025$ and $\sqrt{\Delta'}=55$, so $x=-5+55=50$ (satisfies the condition) or $x=-60$ (rejected).

So the plan required $50$ products per day.

Answer. $50$ products.

Let the speed of the boat in still water be $x$ (km/h), $x>2$. From the problem we have the equation

$$\frac{35}{x+2}+\frac{35}{x-2}=6\Leftrightarrow35(x-2)+35(x+2)=6(x^2-4)\Leftrightarrow3x^2-35x-12=0.$$

We have $\Delta=1225+144=1369$ and $\sqrt\Delta=37$, so $x=\dfrac{35+37}{6}=12$ (satisfies the condition) or $x=\dfrac{35-37}{6}=-\dfrac13$ (rejected).

So the speed of the boat in still water is $12$ km/h.

Answer. $12$ km/h.

Let the rate of each reduction be $r$ ($0<r<1$). From the problem we have the equation

$$5(1-r)^2=4.05\Leftrightarrow(1-r)^2=0.81\Leftrightarrow1-r=0.9\ \text{(since }1-r>0).$$

Hence $r=0.1$.

So the price is reduced by $10\%$ each time.

Answer. $10\%$.

Let the width of the border be $x$ (cm), $x>0$. We have the equation

$$(12+2x)(8+2x)=140\Leftrightarrow4x^2+40x-44=0\Leftrightarrow x^2+10x-11=0.$$

We have $a+b+c=1+10-11=0$, so $x_1=1$ (satisfies the condition) and $x_2=-11$ (rejected).

So the border is $1$ cm wide.

Answer. $1$ cm.

a) We have $\Delta'=1+35=36$, so $x=1+6=7$ or $x=1-6=-5$.

b) Let the original number of students be $x$ ($x\in\mathbb{N}^*$). We work in thousands of dong. From the problem we have the equation

$$\frac{4200}{x}-\frac{4200}{x+2}=10\Leftrightarrow8400=10x(x+2)\Leftrightarrow x^2+2x-840=0.$$

We have $\Delta'=1+840=841$ and $\sqrt{\Delta'}=29$, so $x=-1+29=28$ (satisfies the condition) or $x=-30$ (rejected).

c) In fact there are $30$ students, and each pays $4200\div30=140$ (thousand dong).

So a) $x=7$, $x=-5$; b) there were $28$ students at first; c) each student pays $140$ thousand dong.

Answer. a) $7$; $-5$. b) $28$ students. c) $140$ thousand dong.

Let the number of $5$ thousand dong cuts be $x$ ($x\in\mathbb{N}$, $x<12$). Then the price is $60-5x$ (thousand dong) and the number of tickets sold is $300+50x$. From the problem we have the equation

$$(60-5x)(300+50x)=20\,000\Leftrightarrow250x^2-1500x+2000=0\Leftrightarrow x^2-6x+8=0.$$

We have $\Delta'=9-8=1$, so $x=4$ or $x=2$ (both satisfy the condition). With $x=2$: price $50$ thousand dong, $400$ tickets; with $x=4$: price $40$ thousand dong, $500$ tickets. The cinema chooses the option with more tickets, so $x=4$.

So the new ticket price is $40$ thousand dong.

Remark. With $x=3$ (price $45$ thousand dong, $450$ tickets), the revenue is $20.25$ million dong, which is higher; the quadratic functions of grade 10 will help find the best price.

Answer. $40$ thousand dong.