From Förberedande kurs i matematik 1

Revision as of 08:44, 9 September 2008 (edit) Tekbot (Talk | contribs) m (Lösning 2.2:2c moved to Solution 2.2:2c: Robot: moved page) ← Previous diff		Current revision (14:17, 23 September 2008) (edit) (undo) Tek (Talk | contribs) m
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-	{{NAVCONTENT_START}}	+	We can simplify the left-hand side in the equation by expanding the squares using the squaring rule
-	<center> [[Image:2_2_2c.gif]] </center>	+
-	{{NAVCONTENT_STOP}}	+	{{Displayed math||<math>\begin{align}
		+	(x+3)^{2}-(x-5)^{2}
		+	&= (x^{2}+2\cdot 3x+3^{2})-(x^{2}-2\cdot 5x+5^{2})\\[5pt]
		+	&= x^{2}+6x+9-x^{2}+10x-25\\[5pt]
		+	&=16x-16\,\textrm{.}
		+	\end{align}</math>}}
		+
		+	Thus, the equation is
		+
		+	{{Displayed math||<math>16x-16=6x+4\,\textrm{.}</math>}}
		+
		+	Now, move all ''x'''s to the left-hand side (subtract 6''x'' from both sides) and the constants to the right-hand side (add 16 to both sides)
		+
		+	{{Displayed math||<math>\begin{align}
		+	16x-6x&=4+16\,,\\[5pt]
		+	10x&=20\,\textrm{.}
		+	\end{align}</math>}}
		+
		+	Divide both sides by 10 to get the answer
		+
		+	{{Displayed math||<math>x=\frac{20}{10}=2\,\textrm{.}</math>}}
		+
		+	Finally, we check that <math>x=2</math> satisfies the equation in the exercise
		+
		+	{{Displayed math||<math>\begin{align}
		+	\text{LHS} &= (2+3)^{2}-(2-5)^{2} = 5^{2}-(-3)^{2} = 25-9 = 16,\\[5pt]
		+	\text{RHS} &= 6\cdot 2+4 = 12+4 = 16\,\textrm{.}
		+	\end{align}</math>}}

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Current revision

We can simplify the left-hand side in the equation by expanding the squares using the squaring rule

\displaystyle \begin{align} (x+3)^{2}-(x-5)^{2} &= (x^{2}+2\cdot 3x+3^{2})-(x^{2}-2\cdot 5x+5^{2})\\[5pt] &= x^{2}+6x+9-x^{2}+10x-25\\[5pt] &=16x-16\,\textrm{.} \end{align}

Thus, the equation is

\displaystyle 16x-16=6x+4\,\textrm{.}

Now, move all x's to the left-hand side (subtract 6x from both sides) and the constants to the right-hand side (add 16 to both sides)

\displaystyle \begin{align} 16x-6x&=4+16\,,\\[5pt] 10x&=20\,\textrm{.} \end{align}

Divide both sides by 10 to get the answer

\displaystyle x=\frac{20}{10}=2\,\textrm{.}

Finally, we check that \displaystyle x=2 satisfies the equation in the exercise

\displaystyle \begin{align} \text{LHS} &= (2+3)^{2}-(2-5)^{2} = 5^{2}-(-3)^{2} = 25-9 = 16,\\[5pt] \text{RHS} &= 6\cdot 2+4 = 12+4 = 16\,\textrm{.} \end{align}