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.

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.

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Thành Đạt 5 months 2021-08-15T00:42:02+00:00 2 Answers 27 views 0

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    0
    2021-08-15T00:43:07+00:00

    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

    0
    2021-08-15T00:43:51+00:00

    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

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