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The product of the numbers that are 3 and 7 more than a positive number is 600. What is the original number?

Answer and explanation

Correct answer: \(-5+2\sqrt{151}\)

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.

Related tags

Quadratic EquationsWord ProblemsQuadratic FormulaNumber ProblemsAlgebraic EquationsWord Problems And ApplicationsMathematicsClass 10 Mcq

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