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Learn how to solve equations with extraneous solutions and apply the concept to solve real-world problems. Includes step-by-step examples and guided practice.
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Solve 6 – x=x. 6 – x=x 2 2 ( 6 – x ) =x 2 0 = x + x – 6 6 – x = x 2 EXAMPLE 4 Solve an equation with an extraneous solution SOLUTION Write original equation. Square each side. Simplify. Write in standard form. 0 = (x – 2)(x + 3) Factor. x – 2 = 0 orx + 3 = 0 Zero-product property x = 2 orx = – 3 Solve for x.
2 = 2 3 = –3 ? Ifx = 2:6 – 2 = 2 ANSWER ? Because –3 does not check in the original equation, it is an extraneous solution. The only solution of the equation is 2. Ifx = –3: 6 –(–3) = –3 EXAMPLE 4 Solve an equation with an extraneous solution CHECKCheck 2 and –3 in the original equation.
The hull speed s(in nautical miles per hour) of a sailboat can be estimated using the formula s =1.34 lwhere l is the length (in feet) of the sailboat’s waterline, as shown. Find the length (to the nearest foot) of the sailboat’s waterline if it has a hull speed of 8 nautical miles per hour. EXAMPLE 5 Solve a real-world problem SAILING
s = 1.34 l 8 8 1.34 1.34 8 = 1.34 l =l 2 ANSWER 2 = ( l ) 35.6 l The sailboat has a waterline length of about 36 feet. EXAMPLE 5 Solve a real-world problem SOLUTION Write original equation. Substitute8fors. Divide each side by1.34. Square each side. Simplify.
ANSWER 4 5. Solve3 x + 4 = x. ANSWER 24feet for Examples 4 and 5 GUIDED PRACTICE 6. WHAT IF?In Example 5, suppose the sailboat’s hull speed is 6.5 nautical miles per hour. Find the sailboat’s waterline length to the nearest foot.