Solve application problems using quadratic equations.

Question

A positive real number is 6 less than another. If the sum of the squares of the two numbers is 38, then find the numbers.

Answer 1

Translate the problems into equations

Let x = one number, y= the other

x=y-6 or y=x+6

sum of squares =38 or x² + y² = 38

what you might notice is these equations graphically are a straight line and a circle, which intersect either 0, 1 or 2 times. There may be zero, one or two possible solutions

You could just graph them carefully and see where they intersect.

Or do it algebraically.

x²+y² = 38 substitute from the 1st equation x=y-6 to get

(y-6)²+y²=38

Expand and simplify

y²-12y+36+y²=38

2y²-12y-2+0

divide by 2

y²-6y-1=0

You could use the quadratic formula, complete the square, or factor if possible

Take one half the coefficient of the y term and square it, getting (6/2)² =3² = 9

Add and subtract 9, to complete the square:

y²-6y+9 -1-9 = 0

(y-3)² = 10

y-3 = plus or minus the square root of 10. Square root of 10 is approximately 3.16

y=3 plus 3.16 or 3 minus 3.16

y=6.16 or -0.16

x=y-6 = 6.16-6 or -0.16-6 = 0.16 or -6.16

Since the problem says the real number is positive, we can reject the negative solution leaving

x=0.16 and y=6.16 approximately, rounded off to two decimal places

Or use the quadratic formula getting a=1, b=-6, c=-1 and substitute y=(6 plus (36+4)^1/2)/2

y=3+(1/2)(40^1/2)=3+10^1/2 then x=y-6=-3+10^1/2-6 = 6.16-6=0.16 x=0.16, y=6.16

Check the answer

x is 6 less than y 0.16=6.16-6

also square them both and sum them to get approximately 38

Graphically, the quadratic equation is a circle with center at the origin and radius = (38)^1/2 = a little more than 6.16

the x=y-6 or y=x+6 equation is a straight line with slope = 1 or 45 degrees and intersecting the y axis at y=6

If you can draw this or visualize it, the line crosses the circle twice, once in the 1st quadrant, near the top of the circle, with positive values for x and y, and crosses the circle also in the 3rd quadrant with negative values for x and y near the far left side of the circle. Ignore the negative values.