Your intention with the wording "consecutive integers of 12 feet" is not clear. Shown are triangles with sides 12 and 13 (a nice Pythagorean triple) as well as 11 and 12.

7 0

3 years ago

Diego is solving the equation x^2-12x = 21

uysha [10]

Answer:

The solutions to the quadratic equations will be:

$x=\sqrt{57}+6,\:x=-\sqrt{57}+6$

Step-by-step explanation:

Given the expression

$x^2-12x\:=\:21$

Let us solve the equation by completing the square

$x^2-12x\:=\:21$

Add (-6)² to both sides

$x^2-12x+\left(-6\right)^2=21+\left(-6\right)^2$

simplify

$x^2-12x+\left(-6\right)^2=57$

Apply perfect square formula: (a-b)² = a²-2ab+b²

i.e.

$x^2-12x+\left(-6\right)^2=\left(x-6\right)^2$

so the expression becomes

$\left(x-6\right)^2=57$

$\mathrm{For\:}f^2\left(x\right)=a\mathrm{\:the\:solutions\:are\:}f\left(x\right)=\sqrt{a},\:-\sqrt{a}$

solve

$x-6=\sqrt{57}$

add 6 to both sides

$x-6+6=\sqrt{57}+6$

Simplify

$x=\sqrt{57}+6$

also solving

$x-6=-\sqrt{57}$

add 6 to both sides

$x-6+6=-\sqrt{57}+6$

Simplify

$x=-\sqrt{57}+6$

Therefore, the solutions to the quadratic equation will be:

$x=\sqrt{57}+6,\:x=-\sqrt{57}+6$

5 0

2 years ago