college algebra word problems.

Question

The length of each side of a square is 3 in. more than the length of each side of a smaller square. The sum of the ares of the squares is 149 in squared. Find the lengths of the sides of the two squares.

Answer 1

Let us say the side of the small square is a. Side of the large square (say b) is 3 in. more than a. 

 

b = a+3

 

Area of small square = a² in²

 

Area of large square = b² = (a+3)² in²

 

Sum of the areas = 149 in²

 

a² + (a+3)² = a² + (a²+2.a.3+3²) = 2a²+6a+9 = 149 in^2.

 

2a²+6a-140 = 0 ... It turns out to be a²+3a-70=0. Product of roots is -70 and sum of roots is -3. Some can quickly think of 7 and -10. Or some can observe what another solution provider Arthur D has observed. 

 

Solving the above quadratic equation will give solutions for a.

 

Solving gives a=-10 and a=7. Sides cannot be negative. So a has to be 7.

 

Small side = 7 in. and large side = 7+3 = 10 in.

 

Hope this helps.

 

 

 

 

 

Answer 2

algebraically...

(x+3)^2+x^2=149

x^2+6x+9+x^2=149

2x^2+6x+9-149=0

2x^2+6x-140=0

x^2+3x-70=0

(x+10)(x-7)=0

x-7=0

x=7 inches

x+3=10 inches

 

just an observation...

149 is the sum of the two squares 100 and 49

√100=10 and √49=7