Solution 2.1:2c
From Förberedande kurs i matematik 1
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- | + | We obtain the answer by using the squaring rule, <math>(a+b)^2=a^2+2ab+b^2,</math> on the quadratic term and expanding the other bracketed terms | |
- | < | + | |
- | {{ | + | {{Displayed math||<math>\begin{align} |
+ | (3x+4)^2&-(3x-2)(3x-8)\\ | ||
+ | &=\big( (3x)^2+2\cdot 3x \cdot 4 +4^2 \big) - (3x\cdot 3x-3x\cdot 8 - 2\cdot 3x+ 2\cdot 8)\\ | ||
+ | &= (9x^2+24x+16)-(9x^2-24x-6x+16)\\ | ||
+ | &=(9x^2+24x+16)-(9x^2-30x+16)\\ | ||
+ | &=(9x^2+24x+16)-9x^2+30x-16\\ | ||
+ | &=9x^2-9x^2+24x+30x+16-16\\ | ||
+ | &=0+54x+0\\ | ||
+ | &= 54x\,\textrm{.} | ||
+ | \end{align}</math>}} |
Current revision
We obtain the answer by using the squaring rule, \displaystyle (a+b)^2=a^2+2ab+b^2, on the quadratic term and expanding the other bracketed terms
\displaystyle \begin{align}
(3x+4)^2&-(3x-2)(3x-8)\\ &=\big( (3x)^2+2\cdot 3x \cdot 4 +4^2 \big) - (3x\cdot 3x-3x\cdot 8 - 2\cdot 3x+ 2\cdot 8)\\ &= (9x^2+24x+16)-(9x^2-24x-6x+16)\\ &=(9x^2+24x+16)-(9x^2-30x+16)\\ &=(9x^2+24x+16)-9x^2+30x-16\\ &=9x^2-9x^2+24x+30x+16-16\\ &=0+54x+0\\ &= 54x\,\textrm{.} \end{align} |