Word Problems and Applications

Using distance = speed × time, we get \(x(x+5)=84\). Thus, \(x^2+5x-84=0\), which factors as \((x+12)(x-7)=0\). Therefore, \(x=-12\) or \(x=7\). Since time cannot be negative, the valid answer is 7 hours. Exam tip: Reject any negative root when the variable represents time or distance.

Total work is (x(x+3)=180), giving (x=12). For total work, multiply days and work per day.

The total number of questions is x(x+4). Hence, x(x+4)=320, giving x²+4x−320=0. Factoring, (x+20)(x−16)=0, so x=−20 or x=16. Since the number of days cannot be negative, x=16 is valid. Exam tip: In word problems, calculate both roots of the quadratic and retain only the value that fits the real-world condition.

The total number of books gives the equation \(x(x+9)=682\). Thus, \(x^2+9x-682=0\), which factors as \((x-22)(x+31)=0\). Hence, \(x=22\) or \(x=-31\). Since the number of shelves cannot be negative, the valid answer is 22. Exam tip: In word problems, retain only the positive root that fits the real-life context.

The total-chair condition gives \(x(x+15)=1000\), so \(x^2+15x-1000=0\). Factoring, \((x-25)(x+40)=0\), which gives \(x=25\) or \(x=-40\). Since the number of rows cannot be negative, the valid answer is 25. In such word problems, always reject the negative root on the basis of the context.

The total number of plants gives the equation \(x(x-6)=667\), or \(x^2-6x-667=0\). Factoring, \((x-29)(x+23)=0\), so \(x=29\) or \(x=-23\). Since the number of rows cannot be negative, \(x=29\) is the valid answer. Exam tip: In word problems, retain only the root that is meaningful in the given context.

The total number of questions gives \(x(x-5)=414\). Thus, \(x^2-5x-414=0\), which factors as \((x-23)(x+18)=0\). Therefore, \(x=23\) or \(x=-18\). Since the number of students cannot be negative, the valid answer is \(23\). Exam tip: in word problems, reject algebraic roots that are impossible in the given context.

Let the original number be \(x\). Then \((x+3)(x+7)=600\), which gives \(x^2+10x-579=0\). Using the quadratic formula, \(x=-5\pm2\sqrt{151}\). Since the number is positive, the valid value is \(x=-5+2\sqrt{151}\approx14.6\). Option 15 is a close distractor, but it gives \(18\times22=396\), not 600. Exam tip: Form the equation first and then use the stated condition to select the valid root.

Let the original number be \(x\). Then \((x-5)(x+8)=464\), which simplifies to \(x^2+3x-504=0\), or \((x-21)(x+24)=0\). Thus, \(x=21\) or \(x=-24\); the positive original number is 21. In such problems, carefully translate ‘5 less’ as \(x-5\) and ‘8 more’ as \(x+8\).

Let the original number be x. Then \((x-2)(x+9)=620\), so \(x^2+7x-18=620\), or \(x^2+7x-638=0\). Factoring gives \((x-22)(x+29)=0\). Thus x=22 or x=-29; since the number is specified as positive, the correct answer is 22. Exam tip: verify the result in the original condition: \((22-2)(22+9)=20\times31=620\).

Let the number be \(x\). Then \(x^2=14x+240\), so \(x^2-14x-240=0\). Factoring gives \((x-24)(x+10)=0\), hence \(x=24\) or \(x=-10\). Since the question asks for the positive number, the answer is 24. Check: \(24^2=576\) and \(14\times24+240=576\). Exam tip: Form the quadratic equation directly from the statement, find both roots, and then apply the condition given in the question.

Let the number be \(x\). Then \(x^2-16x=425\), so \(x^2-16x-425=0\). Adding 64 to both sides gives \((x-8)^2=489\), hence \(x=8\pm\sqrt{489}\). Since \(8-\sqrt{489}<0\), the positive number is \(8+\sqrt{489}\). Exam tip: substitute the selected value back into the original equation; a quadratic word problem need not have an integer answer.

Let the number be \(x\). Then \(x^2+x=930\), so \(x^2+x-930=0\). Factoring gives \((x-30)(x+31)=0\), hence \(x=30\) or \(x=-31\). The positive number is therefore 30. As an exam tip, verify the answer by checking \(30^2+30=900+30=930\); choosing 31 gives 992, not 930.

(x^2+156=1057) gives (x^2=901), not a listed perfect square, so this is a data-error type question. In exams, check consistency before solving.

(x^2-175=550) gives (x^2=725), so no listed option should be correct. In such questions, check data consistency.

((x+4)^2=900) and (x+4=30), so (x=26). If the positive case is clear, take the positive square root.

((x-6)^2=784) and (x-6=28), so (x=34). After taking square root, choose the sign according to the condition.

Let the son's age be x years. Then the father's age is x + 28 years. Thus, x(x + 28) = 1485, or x² + 28x − 1485 = 0. Factoring gives (x − 27)(x + 55) = 0. Since age cannot be negative, x = 27, so the father's age is 27 + 28 = 55 years. Exam tip: In age problems, reject the negative root and verify the positive root in the original conditions.

If the daughter's age is (x), then (x(x+24)=640), giving (x=16). In age problems, ignore the negative root.

From the condition, \((x+5)(x-5)=875\). Using the difference-of-squares identity gives \(x^2-25=875\), so \(x^2=900\) and \(x=\pm 30\). Since age cannot be negative, the present age is 30 years. In age problems, always reject the negative root.