Question

A positive real number is 2 less than another. When 4 times the larger is added to the square of the smaller, the result is 49. Find the numbers

Answer 1

Answer:

The numbers are

$-2+3\sqrt{5}$   and $3\sqrt{5}$

Step-by-step explanation:

Let

x -----> the smaller positive real number

y -----> the larger positive real number

we know that

A positive real number is 2 less than another

so

$x=y-2$

$y=x+2$ ----> equation A

When 4 times the larger is added to the square of the smaller, the result is 49

so

$4y+x^2=49$ ----> equation B

substitute equation A in equation B

$4(x+2)+x^2=49$

solve for x

$x^2+4x-41=0$

The formula to solve a quadratic equation of the form

$ax^{2} +bx+c=0$

is equal to

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

in this problem we have

$x^2+4x-41=0$

so

$a=1\\b=4\\c=-41$

substitute in the formula

$x=\frac{-4\pm\sqrt{4^{2}-4(1)(-41)}} {2(1)}$

$x=\frac{-4\pm\sqrt{180}} {2}$

$x=\frac{-4\pm6\sqrt{5}} {2}$

$x=-2\pm3\sqrt{5}$

so

The positive real number is

$x=-2+3\sqrt{5}$

Find the value of y

$y=x+2$

$y=-2+3\sqrt{5}+2$

$y=3\sqrt{5}$