Word Problems and Applications

Let the breadth be \(x\) cm. Then the length is \(x+4\) cm, so \(x(x+4)=320\), or \(x^2+4x-320=0\). Factoring gives \((x-16)(x+20)=0\), so \(x=16\) or \(x=-20\). Since a dimension cannot be negative, the breadth is 16 cm. Exam tip: In measurement problems, reject any negative root.

Let the breadth be \(x\) cm. Then the length is \(x+6\) cm, so \(x(x+6)=667\), giving \(x^2+6x-667=0\). Factoring, \((x-23)(x+29)=0\), gives \(x=23\) or \(x=-29\). A length or breadth cannot be negative, so the breadth is 23 cm. Option D is the length, not the breadth. Exam tip: verify the answer by multiplying the breadth and the corresponding length to check the given area.

(x+(x+15)=53) gives (x=19). This is a simple linear case, so every word problem is not quadratic.

Option (B) forms (x(x+3)=88), which is quadratic. When a variable is multiplied by a variable expression, an (x^2) term appears.

Area of a rectangle is length (\times) breadth, so (x(x+10)=651) is correct. Identifying the formula is the first step.

The total number of plants gives \(x(x-4)=525\), or \(x^2-4x-525=0\). Factoring, \((x-25)(x+21)=0\), so \(x=25\) or \(x=-21\). Since the number of rows cannot be negative, \(x=25\) is the valid answer. In such word problems, always reject a negative value for a physical quantity.

The equation is (x(x+2)=483), giving (x=21). For total count, multiply students and questions per student.

We have (x(x-30)=175), so (x^2-30x-175=0), which factors as ((x-35)(x+5)=0). Thus, (x=35) or (x=-5). Since an age cannot be negative, the father's age is 35 years. Option B would make the son's age (25-30=-5) years, so it is invalid. Exam tip: After solving an age equation, always reject roots that give a negative or otherwise unrealistic age.

Let the original number be \(x\). Then \((x+2)(x+3)=306\), which gives \(x^2+5x-300=0\). Factoring, \((x+20)(x-15)=0\), so \(x=-20\) or \(x=15\). Since the question specifies a positive number, the original number is \(15\). Exam tip: Form the algebraic equation first and then select the root that satisfies the condition in the question.

Let the original number be \(x\). Then \((x-4)(x+5)=360\), which gives \(x^2+x-380=0\). Factoring, \((x-19)(x+20)=0\), so \(x=19\) or \(x=-20\). Since the question asks for the positive number, the answer is \(19\). In such problems, write “4 less” as \(x-4\) and “5 more” as \(x+5\).

Let the smaller integer be \(x\). Then the larger integer is \(x+1\), so \(x(x+1)=156\), giving \(x^2+x-156=0\). Factoring, \((x-12)(x+13)=0\). Since the integers are positive, \(x=12\), and the larger integer is therefore \(13\). Option B is the smaller integer, not the larger one. Exam tip: For consecutive-integer problems, represent the numbers as \(x\) and \(x+1\) before forming the quadratic equation.

If the smaller odd integer is (x), then (x(x+2)=195). (x=13) is correct because (13\times15=195).

Let the smaller even integer be x. The next consecutive even integer is x + 2. Thus, x(x + 2) = 168, giving x² + 2x − 168 = 0 = (x − 12)(x + 14). Since the integers are positive, x = 12, so the larger integer is 12 + 2 = 14. Exam tip: consecutive even integers always differ by 2.

Let the number be (x), so (x+\frac{1}{x}=\frac{5}{2}). This gives (2x^2-5x+2=0), so the larger number is (2).

Let the breadth be \(x\) units. Then the length is \(x+3\) units, so \(x(x+3)=70\), or \(x^2+3x-70=0\). Factoring gives \((x+10)(x-7)=0\). Since a breadth must be positive, \(x=7\), and the length is \(7+3=10\) units. The root \(-10\) is not physically meaningful. Exam tip: In geometry word problems, reject roots that make a length or breadth negative.

The sides are (x) and (17-x) because (2(l+b)=34). From (x(17-x)=60), the sides are (12) and (5).

Let the breadth be \(x\) units. Then the length is \(x+4\) units. Using the area, \(x(x+4)=96\), so \(x^2+4x-96=0\). Factoring gives \((x-8)(x+12)=0\), yielding \(x=8\) or \(x=-12\). Since a breadth cannot be negative, the correct answer is 8. Exam tip: In geometry word problems, reject any negative root because lengths must be positive.

Let the side of the square be \(x\). Its area is \(x^2\) and its perimeter is \(4x\). Therefore, \(x^2=4x+45\), giving \(x^2-4x-45=0\). Factoring, \((x-9)(x+5)=0\), so \(x=9\) or \(x=-5\). Since a length cannot be negative, the valid answer is 9. Exam tip: For a geometrical length, reject any negative root of the quadratic equation.

If one number is (x), the other is (15-x). From (x(15-x)=54), the numbers are (9) and (6).

If the smaller number is (x), the larger is (x+2). From (x^2+(x+2)^2=130), (x=7), so the larger number is (9).