So \( n(n+2) = 210 \). - United Radiology

["# Solve the Equation: So ( n(n+2) = 210 )", "If you're looking to solve the equation So ( n(n+2) = 210 ), you're tackling a classic quadratic problem. This equation appears in math curriculum and is often used in algebraic problem-solving. Whether you're a student, teacher, or math enthusiast, understanding how to solve equations like this is essential. In this article, we’ll break down step-by-step how to solve ( n(n+2) = 210 ), explore its roots, and offer practical applications.", "---", "## Understanding the Equation", "The equation ( n(n+2) = 210 ) involves a product of two consecutive-like integers (since ( n ) and ( n+2 ) differ by 2), and we want their product to equal 210. Rearranged as a standard quadratic, this equation helps strengthen algebraic reasoning and problem-solving skills.", "---", "## Rearranging to Standard Quadratic Form", "Start by expanding and moving all terms to one side:", "[
\nn(n+2) = 210
\n]
\n[
\nn^2 + 2n = 210
\n]
\n[
\nn^2 + 2n - 210 = 0
\n]", "Now the equation is in standard quadratic form:
\n[
\nn^2 + 2n - 210 = 0
\n]", "---", "## Solving the Quadratic Equation", "We can solve this quadratic using factoring, completing the square, or the quadratic formula. Let’s try factoring since 210 has clear factor pairs.", "Look for two numbers that multiply to (-210) and add to (2):", "- (15 \ imes (-14) = -210) and (15 + (-14) = 1) ❌
\n- (21 \ imes (-10) = -210) and (21 + (-10) = 11) ❌
\n- (30 \ imes (-7) = -210) and (30 + (-7) = 23) ❌
\n- (15 \ imes (-14)) was close — wait: actually, no integer pair adds to +2 direktly? Let's verify carefully.", "Try factoring directly:", "We seek integers ( p ) and ( q ) such that
\n[
\n(n + p)(n + q) = n^2 + (p+q)n + pq = n^2 + 2n -210
\n]
\nSo:
\n- ( p + q = 2 )
\n- ( pq = -210 )", "Try ( p = 15, q = -14 ):
\n( 15 + (-14) = 1 <br/>\ne 2 )
\nTry ( p = 21, q = -10 ): (21 -10 = 11)
\nWait — actually, ( p = 15 ) and ( q = -14 ) gives sum 1. Not working.", "But note: (15 \ imes (-14) = -210) but sum is (1), not 2.", "Try ( p = 10, q = -21 ): sum (-11) ❌
\nWait — correct approach: factor (-210) pairs until sum is 2.", "Try ( 15 \ imes (-14) ) → sum 1
\nBut what about ( 14 \ imes (-15) )? sum -1
\nWait — try ( 30 \ imes (-7) ): sum 23
\nTry ( 21 \ imes (-10) = -210 ), sum = 11
\nTry ( 35 \ imes (-6) = -210 ), sum = 29
\nWait — actually, no simple integer pair adds to 2.", "That suggests we use the quadratic formula:", "[
\nn = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\n]
\nFor ( a = 1, b = 2, c = -210 ):", "[
\nn = \frac{-2 \pm \sqrt{(2)^2 - 4(1)(-210)}}{2(1)} = \frac{-2 \pm \sqrt{4 + 840}}{2} = \frac{-2 \pm \sqrt{844}}{2}
\n]", "Wait — ( \sqrt{844} ) is not a perfect square. Let’s simplify:", "But earlier we wanted integer solutions. Let’s double-check factoring.", "---", "Wait — actually, let’s reverse-engineer the factors.", "We want two numbers that multiply to (-210) and add to (2). Try:", "- (15) and (-14): (15 \ imes -14 = -210), but sum = 1
\n- (21) and (-10): sum = 11
\n- (30) and (-7): sum = 23
\n- (35) and (-6): sum = 29", "But 35 and -6 give sum 29 — not helpful.", "Wait: trying ( n = 14 ):
\n( 14 \ imes (14+2) = 14 \ imes 16 = 224 ) → too big
\nTry ( n = 13 ): (13 \ imes 15 = 195)
\nTry ( n = 14 ): (14 \ imes 16 = 224)
\nSo between 13 and 14 — not integer?", "But wait — let's solve the quadratic properly.", "---", "Correcting Factorization Attempt:", "We solve:
\n[
\nn^2 + 2n - 210 = 0
\n]", "Try factoring:", "Find ( p ) and ( q ) such that:
\n( p + q = 2 ), ( pq = -210 )", "Try ( p = 15 ), ( q = -14 ): product -210, sum +1 ❌
\nTry ( p = -14 ), ( q = 15 ): same
\nTry ( p = 10 ), ( q = -21 ): product -210, sum -11
\nTry ( p = 21 ), ( q = -10 ): sum 11
\nTry ( p = -15 ), ( q = 14 ): sum -1
\nWait — actually, no integer pair adds exactly 2.", "So no perfect factorization — use the quadratic formula.", "[
\nn = \frac{-2 \pm \sqrt{4 + 840}}{2} = \frac{-2 \pm \sqrt{844}}{2}
\n]", "But simplify ( \sqrt{844} ):", "[
\n844 = 4 \ imes 211 \Rightarrow \sqrt{844} = 2\sqrt{211}
\n]", "So:", "[
\nn = \frac{-2 \pm 2\sqrt{211}}{2} = -1 \pm \sqrt{211}
\n]", "Now ( \sqrt{211} \approx 14.52 ), so:", "[
\nn \approx -1 + 14.52 = 13.52 \quad \ ext{or} \quad n \approx -1 - 14.52 = -15.52
\n]", "But wait — this contradicts earlier trials?", "Wait — double-check expansion:", "[
\nn(n+2) = n^2 + 2n = 210 \Rightarrow n^2 + 2n - 210 = 0
\n]", "Yes, correct.", "Now verify discriminant:", "[
\n\Delta = 2^2 - 4(1)(-210) = 4 + 840 = 844
\n]", "( 844 = 4 \ imes 211 ), and 211 is prime → irrational roots.", "Thus, exact solutions are:", "[
\nn = -1 \pm \sqrt{211}
\n]", "Only the positive root is meaningful if ( n ) is real and positive:", "[
\nn = -1 + \sqrt{211} \approx 13.52
\n]", "But the question might expect exact form or perhaps there’s a typo?", "Wait — let’s check if 210 can be written as product of two numbers differing by 2.", "Try ( n = 13 ): (13 \ imes 15 = 195)
\n( n = 14 ): (14 \ imes 16 = 224)
\nSo no integer solution?", "But equation ( n(n+2) = 210 ) implies ( (n+1)^2 - 1 = 210 \Rightarrow (n+1)^2 = 211 \Rightarrow n+1 = \sqrt{211} ) → same result.", "So no integer solution.", "---", "## Final Answer", "The equation ( n(n+2) = 210 ) simplifies to the quadratic equation:", "[
\nn^2 + 2n - 210 = 0
\n]", "Using the quadratic formula:", "[
\nn = \frac{-2 \pm \sqrt{844}}{2} = \frac{-2 \pm 2\sqrt{211}}{2} = -1 \pm \sqrt{211}
\n]", "Thus, the solutions are:", "[
\nn = -1 + \sqrt{211} \quad \ ext{and} \quad n = -1 - \sqrt{211}
\n]", "The positive solution ( n = -1 + \sqrt{211} ) (≈ 13.52) satisfies the equation algebraically.", "---", "## Practical Application", "Equations like ( n(n+2) = C ) model situations where two values differing by 2 multiply to a constant—common in optimization, physics (e.g., motion over intervals), and algebraic modeling.", "For example, suppose two numbers differ by 2 and their product is 210 — what are they? Their average is near ( \sqrt{210} \approx 14.5 ), so ( 13 \ imes 15 = 195 ), ( 14 \ imes 16 = 224 ); 210 lies between — reinforcing no integer pair exists, validating our algebraic result.", "---", "## Summary", "- Rearrange to standard quadratic: ( n^2 + 2n - 210 = 0 )
\n- Use quadratic formula: ( n = \frac{-2 \pm \sqrt{844}}{2} = -1 \pm \sqrt{211} )
\n- No integer solutions, but exact real solutions exist
\n- Ideal for teaching algebraic transformation and quadratic solving", "---", "## Want More?", "- Explore exact vs. approximate solutions
\n- Learn how to factor quadratics with non-integer roots
\n- See real-world problems modeled by similar equations", "---", "SEO Keywords:
\n( n(n+2) = 210 ), solve quadratic equation, algebra quizzes, quadratic formula, integer solutions, real roots, factoring quadratics, Diophantine equations, algebraic problems, solving equations step-by-step", "---", "Understanding equations like ( n(n+2) = 210 ) strengthens logical thinking and mathematical fluency — essential for standardized tests, coding, and real-world problem-solving."]