Solution 2.1:1e
From Förberedande kurs i matematik 1
(Difference between revisions)
(Ny sida: {{NAVCONTENT_START}} <center> Bild:2_1_1e.gif </center> {{NAVCONTENT_STOP}}) |
m |
||
(3 intermediate revisions not shown.) | |||
Line 1: | Line 1: | ||
- | {{ | + | If we use the rule for squaring <math>(a-b)^2 = a^2-2ab+b^2 </math> with <math> a=x </math> and <math> b=7</math>, we obtain directly that |
- | < | + | |
- | {{ | + | {{Displayed math||<math> (x-7)^2=x^2-2 \cdot x \cdot 7 + 7^2 = x^2-14x+49\,\textrm{.}</math>}} |
+ | |||
+ | An alternative is to write the square as <math> (x-7)\cdot (x-7)</math> and then multiply the brackets in two steps | ||
+ | |||
+ | {{Displayed math||<math>\begin{align} | ||
+ | (x-7)\cdot (x-7) &= (x-7)\cdot x - (x-7)\cdot 7 \\[3pt] | ||
+ | &= x\cdot x-7 \cdot x -(x\cdot 7 - 7\cdot 7) \\[3pt] | ||
+ | &= x^2 -7x-(7x-49)\\[3pt] | ||
+ | & \stackrel{*}= x^2-7x-7x+49 \\[3pt] | ||
+ | &= x^2-(7+7)x+49\\[3pt] | ||
+ | &= x^2-14x+49\,\textrm{.} | ||
+ | \end{align}</math>}} | ||
+ | |||
+ | In the line that has been marked with an asterisk, we have removed the bracket and at the same time changed signs on all terms inside the bracket. |
Current revision
If we use the rule for squaring \displaystyle (a-b)^2 = a^2-2ab+b^2 with \displaystyle a=x and \displaystyle b=7, we obtain directly that
\displaystyle (x-7)^2=x^2-2 \cdot x \cdot 7 + 7^2 = x^2-14x+49\,\textrm{.} |
An alternative is to write the square as \displaystyle (x-7)\cdot (x-7) and then multiply the brackets in two steps
\displaystyle \begin{align}
(x-7)\cdot (x-7) &= (x-7)\cdot x - (x-7)\cdot 7 \\[3pt] &= x\cdot x-7 \cdot x -(x\cdot 7 - 7\cdot 7) \\[3pt] &= x^2 -7x-(7x-49)\\[3pt] & \stackrel{*}= x^2-7x-7x+49 \\[3pt] &= x^2-(7+7)x+49\\[3pt] &= x^2-14x+49\,\textrm{.} \end{align} |
In the line that has been marked with an asterisk, we have removed the bracket and at the same time changed signs on all terms inside the bracket.