Word Problems and Applications

Let the smaller integer be \(x\). The next consecutive odd integer is \(x+2\). Thus, \(x^2+(x+2)^2=1570\), which gives \(2x^2+4x-1566=0\) or \(x^2+2x-783=0\). Factoring, \((x-27)(x+29)=0\), so \(x=27\) or \(-29\). Since the integers are positive, the smaller one is 27. Check: \(27^2+29^2=729+841=1570\). Exam tip: consecutive odd integers always differ by 2.

Let the consecutive even numbers be (x) and (x+2). From (x^2+(x+2)^2=1924), (x=30), so the larger number is (32).

If the smaller number is (x) and the larger is (x+13), then (x(x+13)=558), giving (x=18). The larger number is (31).

If the numbers are (x) and (65-x), then (x^2+(65-x)^2=2125). Solving gives (30) and (35).

Taking breadth as (x), (x(x+14)=1152), giving (x=32). For area, multiply length and breadth.

Let the breadth of the rectangle be \(x\) cm. Its length is then \(x+18\) cm, so \(x(x+18)=1240\). Thus, \(x^2+18x-1240=0\), giving \(x=-9\pm\sqrt{1321}\). Since a dimension must be positive, the breadth is \(\sqrt{1321}-9\) cm and the length is \(x+18=9+\sqrt{1321}\) cm. Therefore, option D is correct. In an exam, select only the positive root for a physical dimension; 31 cm is not correct because \(31\times49=1519\), not 1240.

If breadth is (x), then (x^2+(x+10)^2=74^2), giving (x=48). Use Pythagoras for a rectangle with a diagonal.

Let the breadth be \(x\) cm. Then the length is \(x+13\) cm. Using the area, \(x(x+13)=888\), so \(x^2+13x-888=0\). Factoring gives \((x-24)(x+37)=0\). The positive root gives a breadth of 24 cm and a length of 37 cm. Hence, the perimeter is \(2(24+37)=122\) cm. The negative root \(-37\) cannot represent a side length. Exam tip: In rectangle word problems, assign \(x\) to one side and express the other side using the stated difference before forming the quadratic equation.

If one side is (x), the other is (77-x), and (x(77-x)=1480). The solutions are (37) and (40), so the shorter side is (37).

Since the areas are equal, \(x(x+20)=42^2=1764\). Therefore, \(x^2+20x-1764=0\). Using the quadratic formula, \(x=\frac{-20\pm\sqrt{20^2+4(1764)}}{2}=-10\pm2\sqrt{466}\). Because a breadth must be positive, the valid value is \(x=-10+2\sqrt{466}\approx33.17\) m; the negative root is not physically meaningful. In such problems, first form the area equation and then reject any root that makes a length negative.

The area of a square is (side)², so \((x+6)^2=1225=35^2\). Since a side length is positive, \(x+6=35\), giving \(x=35-6=29\). The negative root, \(x+6=-35\), is not valid for a length. Exam tip: Take the positive square root when finding a geometrical length from its area.

The area of a square is (side)², so \((x-8)^2=900\). Thus, \(x-8=\pm 30\), but a side length must be positive; hence \(x-8=30\) and \(x=38\). Option A incorrectly treats 30, the side length, as the value of \(x\). Exam tip: reject a negative root when the expression represents a length.

By Pythagoras, (x^2+(x+18)^2=90^2), giving (x=54). Form the equation by taking the hypotenuse as the largest side.

Let the shorter leg be \(x\) cm. Then the longer leg is \(x+19\) cm. By the Pythagorean theorem, \(x^2+(x+19)^2=109^2\), which simplifies to \(x^2+19x-5760=0\). Its positive solution is \(x=\frac{\sqrt{23401}-19}{2}\approx66.99\) cm, so option A is correct. Option B is incorrect because legs of 60 cm and 79 cm do not produce a hypotenuse of 109 cm. Exam tip: for a length problem, reject the negative root and verify the result using the Pythagorean relation.

From the right triangle, (x^2+60^2=(x+15)^2), giving (x=105). In ladder problems, the ladder is the hypotenuse.

If the base distance is (x), then (x^2+72^2=(x+18)^2), giving (x=135). The hypotenuse is always the greatest length.

Using distance = speed × time, we get \(x(x-65)=750\). Thus, \(x^2-65x-750=0\), which factors as \((x-75)(x+10)=0\). Therefore, \(x=75\) or \(x=-10\). Since speed cannot be negative, the valid answer is 75 km per hour. In the exam, first form the quadratic equation from distance = speed × time, then reject any negative or otherwise physically invalid root.

From (x(x-63)=560), (x=70). The time is (x-63=7) hours, so the correct option is (7) hours.

Downstream speed is (x+3), and (5(x+3)=90), so (x=15). In boat problems, speed changes with direction.

Using distance = speed × time, we get \(x(x+12)=640\). Rearranging gives \(x^2+12x-640=0\), which factors as \((x+32)(x-20)=0\). Thus, \(x=20\) or \(x=-32\); since time cannot be negative, the valid value is 20 hours. Exam tip: For quantities such as time and distance, reject negative roots.