EXAMPLE 3 Standardized Test Practice SOLUTION x 2 – 5x – 36 = 0 Write original equation. (x – 9)(x + 4) = 0 Factor. Zero product property x = 9 or x =

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EXAMPLE 3 Standardized Test Practice SOLUTION x 2 – 5x – 36 = 0 Write original equation. (x – 9)(x + 4) = 0 Factor. Zero product property x = 9 or x = – 4 Solve for x. x – 9 = 0 or x + 4 = 0 ANSWER The correct answer is C.

EXAMPLE 4 Use a quadratic equation as a model Nature Preserve A town has a nature preserve with a rectangular field that measures 600 meters by 400 meters. The town wants to double the area of the field by adding land as shown. Find the new dimensions of the field.

EXAMPLE 4 Use a quadratic equation as a model SOLUTION 480,000 = 240, x + x 2 Multiply using FOIL. 0 = x x – 240,000 Write in standard form. 0 = (x – 200) (x ) Factor. x – 200 = 0x = 0 or Zero product property x = 200 or x = – 1200 Solve for x.

EXAMPLE 4 Use a quadratic equation as a model ANSWER Reject the negative value, – The field’s length and width should each be increased by 200 meters. The new dimensions are 800 meters by 600 meters.

GUIDED PRACTICE for Examples 3 and 4 8. Solve the equation x 2 – x – 42 = 0. SOLUTION x 2 – x – 42 = 0 Write original equation. Factor. Zero product property x = – 6 or x = 7 Solve for x. x + 6 = 0 or x – 7 = 0 (x + 6)(x – 7) = 0

GUIDED PRACTICE for Examples 3 and 4 9. What If ? In Example 4, suppose the field initially measures 1000 meters by 300 meters. Find the new dimensions of the field. SOLUTION New AreaNew Length (meters) New width (meters) = 2(1000)(300) = ( x)(300 + x) = Multiply using FOIL x + 300x + x 2 0 = x x – Write in standard form. 0 = (x – 200) (x ) Factor. x – 200 = 0x = 0 or Zero product property

GUIDED PRACTICE for Examples 3 and 4 x = 200 or x = – 1500 Solve for x. ANSWER Reject the negative value, – The field’s length and width should each be increased by 200 meters. The new dimensions are 1200 meters by 500 meters.