Consider three consecutive positive integers, such that the sum of the

Mathematics

1 answer:

Greeley [361]3 years ago

4 0

Answer:

Step-by-step explanation:

Let x represent the smaller integer. Then we have ...

(x +1)² +(x +2)² = 40x +5

2x² +6x +5 = 40x +5

x² -17x = 0 . . . . . subtract (40x+5), divide by 2

x(x -17) = 0 . . . . . factor

The solution of interest is x = 17.

The smaller integer is 17.

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Which are the solutions of x^2 = –11x + 4?

strojnjashka [21]

This is a quadratic equation, i.e. an equation involving a polynomial of degree 2. To solve them, you must rearrange them first, so that all terms are on the same side, so we get

$x^2 + 11x - 4 = 0$

i.e. now we're looking for the roots of the polynomial. To find them, we can use the following formula:

$x_{1,2} = \frac{-b\pm\sqrt{b^2-4ac}}{2}$

where $x_{1,2}$ is a compact way to indicate both solutions $x_1$ and $x_2$ , while $a,b,c$ are the coefficients of the quadratic equation, i.e. we consider the polynomial $ax^2+bx+c$ .

So, in your case, we have $a=1,\ \ b=11,\ \ c=-4$

Plug those values into the formula to get

$x_{1,2} = \frac{-11\pm\sqrt{121+16}}{2} = \frac{-11\pm\sqrt{137}}{2}$

So, the two solutions are

$x_1 = \frac{-11+\sqrt{137}}{2}$

$x_2 = \frac{-11-\sqrt{137}}{2}$