Question

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

Answer 1

Answer:

There seems to be a typo error but still this has solution.

Let the number be = x

A positive real number is 6 less than another. Means the second will be $(x+6)$

The sum of the square of the two numbers is 38.

$x^{2} +(x+6)^{2} =38$  

=> $x^{2} +36+12x+x^{2} =38$

=> $2x^{2}+12x+36 =38$

=> $2x^{2}+12x=2$

=> $2x^{2}+12x-2=0$

Taking out 2 common;

=> $x^{2}+6x-1=0$

Solving this quadratic equation, we get;

$x=-3+\sqrt{10}$ and $x=-3-\sqrt{10}$

As positive number is needed, we have $-3+\sqrt{10}$

or x = $\sqrt{10}-3$

x = 0.162277

And other number is 6.162277.

Answer 2

• -3 + √10 = 0.162, and
• (-3 + √10) + 6 ⇒ 3 + √10 = 6.162.

Further explanation

Given:

• A positive real number is 6 less than another.
• The sum of the square of the two numbers is 38.

Question:

Find out the numbers.

Problem-solving:

Let us use $\boxed{ \ x \ }$ to represent the positive real number.

This positive real number is 6 less than another. The other number is $\boxed{ \ x + 6 \ }$

The sum of the square of the two numbers is 38, i.e., $\boxed{ \ x^2 + (x + 6)^2 = 38 \ }$

Let us work on expanding it.

$\boxed{ \ x^2 + (x + 6)^2 = 38 \ }$

Recall that $\boxed{ \ (a + b)^2 = a^2 + 2ab + b^2 \ }$

$\boxed{ \ x^2 + x^2 + 12x + 36 = 38 \ }$

$\boxed{ \ 2x^2 + 12x + 36 - 38 = 0 \ }$

$\boxed{ \ 2x^2 + 12x - 2 = 0 \ }$

Both sides are divided by two.

$\boxed{ \ x^2 + 6x - 1 = 0 \ }$

Let us solve the equation by using the quadratic formula.

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

• a = 1
• b = 6
• c = -1

$\boxed{ \ x = \frac{-6 \pm \sqrt{6^2 - 4(1)(-1)}}{2(1)} \ }$

$\boxed{ \ x = \frac{-6 \pm \sqrt{36 + 4}}{2} \ }$

$\boxed{ \ x = \frac{-6 \pm \sqrt{40}}{2} \ }$

$\boxed{ \ x = \frac{-6 \pm 2\sqrt{10}}{2} \ }$

$\boxed{ \ x = \frac{-6}{2} \pm \frac{2\sqrt{10}}{2} \ }$

$\boxed{ \ x = -3 \pm \sqrt{10} \ }$

$\boxed{ \ x_1 = -3 + \sqrt{10} \ . \ . \ . \ x_1 > 10 \ }$ this is fulfilled.

$\boxed{ \ x_2 = -3 - \sqrt{10} \ . \ . \ . \ x_2 < 10 \ }$

• Thus, the positive real number is $\boxed{ \ x = -3 + \sqrt{10} \ }$
• The other number is $\boxed{ \ (-3 + \sqrt{10}) + 6 \rightarrow \boxed{ \ 3 + \sqrt{10} \ }}$

$\boxed{ \sqrt {10} = 3.162 \ }$

• Therefore $\boxed{ \ x = -3 + 3.162 \rightarrow \boxed{ \ 0.162 \ }}$
• And the other number is $\boxed{ \ 0.162 + 6 \rightarrow \boxed{ \ 6.162 \ }}$

Notes:

$\boxed{ \ ax^2 + bx + c = 0, \ \ \ \ \  a \neq 0 \ }$ is a quadratic equation in standard form.

• x is a variable
• a, b, and c are real number coefficients

Methods of solving quadratic equations include:

• Factoring and using the zero product property: $\boxed{ \ (x - a)(x - b) = 0 \ } \ if \ and \ only \ if \ \boxed{ \ (x - a) = 0 \ or \ (x - b) = 0 \ }$
• Using the square root property: $\boxed{ \ x^2 = a, \ then \ x = \pm \sqrt{a} \ }$
• Completing the square: $\boxed{ \ x^2 + bx + \bigg({\frac{b}{2}} \bigg)^2 = \bigg( x + \frac{b}{2} \bigg)^2 \ }$
• Using the quadratic formula: $\boxed{ \ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \ }$

Learn more

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Keywords: a positive real number is 6, less than another, if the sum of the square, 38, then finds, quadratic equation, methods, factoring, the zero product property, the square root, completing, the quadratic formula