Algebra · real student question

Solve for x: x(-5x + 250) = 2625.

Question

Solve for xx:

x(5x+250)=2625x(-5x+250)=2625

Step-by-step solution

  1. Recognise the shape. A product like x(5x+250)x(-5x+250) is the classic revenue or area model: xx units at a price that falls as xx rises. Setting it equal to 26252625 asks which two quantities produce that target — expect two answers, one on each side of the maximum.

  2. Expand and move everything to one side.

    5x2+250x=26255x2+250x2625=0-5x^2+250x=2625\qquad\Longrightarrow\qquad -5x^2+250x-2625=0

  3. Divide by -5 to make the leading coefficient 1. This is the step that turns ugly numbers into factorable ones. Dividing an equation (not an inequality) by a negative is harmless:

    x250x+525=0x^2-50x+525=0

    Note 250/(5)=50250/(-5)=-50 and 2625/(5)=+525-2625/(-5)=+525 — both signs flip.

  4. Factor the monic quadratic. Look for two numbers multiplying to +525+525 and summing to 50-50; both must be negative. Since 525=3527525=3\cdot5^2\cdot7, the candidate pairs are (1,525),(3,175),(5,105),(7,75),(15,35),(21,25)(1,525),(3,175),(5,105),(7,75),(15,35),(21,25), and 15+35=5015+35=50:

    x250x+525=(x15)(x35)x^2-50x+525=(x-15)(x-35)

  5. Apply the zero-product property.

    x15=0x=15,x35=0x=35x-15=0\Rightarrow x=15,\qquad x-35=0\Rightarrow x=35

  6. Verify both in the original equation. At x=15x=15: 15(75+250)=15×175=262515(-75+250)=15\times175=2625 ✓. At x=35x=35: 35(175+250)=35×75=262535(-175+250)=35\times75=2625 ✓. As a structural check, the roots sum to 50=b/a50=-b/a and multiply to 525=c/a525=c/a ✓, and their midpoint 2525 is where the parabola peaks at 25×125=312525\times125=3125 — comfortably above 26252625, as it must be for two solutions to exist.

Answer

x=15orx=35x=15\quad\text{or}\quad x=35

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