Question

A positive real number is 4 less than another. If the sum of the squares of the two numbers is 72, then find the numbers.

Answer 1

Answer:

Our two numbers are:

$2+4\sqrt{2} \text{ and } 4\sqrt{2}-2$

Or, approximately 7.66 and 3.66.

Step-by-step explanation:

Let the two numbers be a and b.

One positive real number is four less than another. So, we can write that:

$b=a-4$

The sum of the squares of the two numbers is 72. Therefore:

$a^2+b^2=72$

Substitute:

$a^2+(a-4)^2=72$

Solve for a. Expand:

$a^2+(a^2-8a+16)=72$

Simplify:

$2a^2-8a+16=72$

Divide both sides by two:

$a^2-4a+8=36$

Subtract 36 from both sides:

$a^2-4a-28=0$

The equation isn't factorable. So, we can use the quadratic formula:

$\displaystyle x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

In this case, a = 1, b = -4, and c = -28. Substitute:

$\displaystyle x=\frac{-(-4)\pm\sqrt{(-4)^2-4(1)(-28)}}{2(1)}$

Evaluate:

$\displaystyle x=\frac{4\pm\sqrt{128}}{2}=\frac{4\pm8\sqrt{2}}{2}=2\pm4\sqrt{2}$

So, our two solutions are:

$\displaystyle x_1=2+4\sqrt{2}\approx 7.66\text{ or } x_2=2-4\sqrt{2}\approx-3.66$

Since the two numbers are positive, we can ignore the second solution.

So, our first number is:

$a=2+4\sqrt{2}$

And since the second number is four less, our second number is:

$b=(2+4\sqrt{2})-4=4\sqrt{2}-2\approx 3.66$

Answer 2

Answer:

$2+4\sqrt{2}\text{ and }4\sqrt{2}-2$

Step-by-step explanation:

Let the large number be $x$. We can represent the smaller number with $x-4$. Since their squares add up to 72, we have the following equation:

$x^2+(x-4)^2=72$

Expand $(x-4)^2$ using the property $(a-b)^2=a^2-2ab+b^2$:

$x^2+x^2-2(4)(x)+16=72$

Combine like terms:

$2x^2-8x+16=72$

Subtract 72 from both sides:

$2x^2-8x-56=0$

Use the quadratic formula to find solutions for $x$:

$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$ for $ax^2+bx+c$

In $2x^2-8x-56$, assign:

• $a\implies 2$
• $b \implies -8$
• $c\implies -56$

Solving, we get:

$x=\frac{-(-8)\pm \sqrt{(-8)^2-4(2)(-56)}}{2(2)},\\x=\frac{8\pm 16\sqrt{2}}{4},\\\begin{cases}x=\frac{8+16\sqrt{2}}{4}, x=\boxed{2+4\sqrt{2}} \\x=\frac{8-16\sqrt{2}}{4}, x=\boxed{2-4\sqrt{2}}\end{cases}$

Since the question stipulates that $x$ is positive, we have $x=\boxed{2+4\sqrt{2}}$. Therefore, the two numbers are $2+4\sqrt{2}$ and $4\sqrt{2}-2$.

Verify:

$(2+4\sqrt{2})^2+(4\sqrt{2}-2)^2=72\:\checkmark$