Mathematics, 28.06.2021 01:00 hardwick744

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.

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Answer from: itsme123427

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 from: hardworker9437

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

Step-by-step explanation:

Let the large number be . 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 :

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

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

Answer from: Quest

3. parallelogram

step-by-step explanation:

we turn each of our sides into direction vectors

ab=b-a=(3 - 7, 1 - 0) = (-4, 1)

bc=c-b=(1,6)

cd=d-c=(4,-1)

da=a-d=(-1,-6)

we see how the opposite sides have negated direction vectors; this is characteristic of a parallelogram. we have congruent sides ab=cd and bc=da with unequal lengths because 1^2+6^2 is different from (-4)^2+1^2. so it's not a square or a rhombus.

we check the dot product of direction vectors of adjacent sides. we have

ab dot bc = -4(1) + 1(6) = 2

that's not zero so the sides are not perpendicular. we've ruled out rectangle.

the direction vectors indicate two pairs of congruent parallel sides, so we choose parallelogram over trapezoid.

Answer from: Quest

answer: all the values are within because it has a pattern that repeats from -6 to 4.

good luck!

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