Solution 2.3:4b

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A first-degree equation which
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A first-degree equation which has <math>x=1+\sqrt{3}</math> as a root is <math>x-(1+\sqrt{3}\,)=0</math>, which we can also write as <math>x-1-\sqrt{3} = 0</math>. In the same way, we have that <math>x-(1-\sqrt{3}\,)=0</math>, i.e., <math>x-1+\sqrt{3}=0</math> is a first-degree equation that has <math>x=1-\sqrt{3}</math> as a root. If we multiply these two first-degree equations together, we get a second-degree equation with <math>x=1+\sqrt{3}</math> and <math>x=1-\sqrt{3}</math> as roots,
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<math>x=\text{ 1}+\sqrt{\text{3}}</math>
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as a root is
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<math>x-\left( \text{1}+\sqrt{\text{3}} \right)=0</math>, which we can also write as
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<math>x-\text{1-}\sqrt{\text{3}}=0</math>. In the same way, we have that
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{{Displayed math||<math>(x-1-\sqrt{3}\,)(x-1+\sqrt{3}\,) = 0\,\textrm{.}</math>}}
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<math>x-\left( \text{1-}\sqrt{\text{3}} \right)=0</math>, i.e.,
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<math>x-\text{1+}\sqrt{\text{3}}=0</math>
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is a first-degree equation that has
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<math>x=\text{ 1-}\sqrt{\text{3}}</math>
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as a root. If we multiply these two first-degree equations together, we get a second-degree equation with
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<math>x=\text{ 1}+\sqrt{\text{3}}</math>
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and
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<math>x=\text{ 1-}\sqrt{\text{3}}</math>. as roots:
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The first factor become zero when <math>x=1+\sqrt{3}</math> and the second factor becomes zero when <math>x=1-\sqrt{3}\,</math>.
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<math>\left( x-\text{1-}\sqrt{\text{3}} \right)\left( x-\text{1+}\sqrt{\text{3}} \right)=0</math>
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Nothing really prevents us from answering with <math>(x-1-\sqrt{3}\,)(x-1+\sqrt{3}\,) = 0</math>, but if we want to give the equation in standard form, we need to expand the left-hand side,
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{{Displayed math||<math>\begin{align}
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(x-1-\sqrt{3}\,)(x-1+\sqrt{3}\,)
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&= x^{2} - x + \sqrt{3}x - x + 1 - \sqrt{3} - \sqrt{3}x + \sqrt{3} - (\sqrt{3}\,)^{2}\\[5pt]
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&= x^{2} + (-x+\sqrt{3}x-x-\sqrt{3}x) + (1-\sqrt{3}+\sqrt{3}-3)\\[5pt]
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&= x^{2}-2x-2
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\end{align}</math>}}
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The first factor become zero when
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to get the equation <math>x^{2}-2x-2=0\,</math>.
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<math>x=\text{ 1}+\sqrt{\text{3}}</math>
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and the second factor becomes zero when
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<math>x=\text{ 1-}\sqrt{\text{3}}</math>.
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Nothing really prevents us from answering with
 
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<math>\left( x-\text{1-}\sqrt{\text{3}} \right)\left( x-\text{1+}\sqrt{\text{3}} \right)=0</math>, but if we want to give the equation in standard form, we need to expand the left-hand side,
 
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Note: Exactly as in exercise a, we can multiply the equation by a non-zero constant ''a''
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<math>\begin{align}
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{{Displayed math||<math>ax^{2}-2ax-2a=0</math>}}
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& \left( x-\text{1-}\sqrt{\text{3}} \right)\left( x-\text{1+}\sqrt{\text{3}} \right)=x^{2}-x+\sqrt{3}-x+1-\sqrt{3}-\sqrt{3}x+\sqrt{3}-\left( \sqrt{3} \right)^{2} \\
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& =x^{2}+\left( -x+\sqrt{3}x-x-\sqrt{3}x \right)+\left( 1-\sqrt{3}+\sqrt{3}-3 \right) \\
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& =x^{2}-2x-2 \\
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\end{align}</math>
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to get the equation
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<math>x^{2}-2x-2=0</math>
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NOTE: Exactly as in exercise
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<math>\text{a}</math>, we can multiply the equation by a non-zero constant
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<math>a</math>,
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<math>ax^{2}-2ax-2a=0</math>
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and still have a second-degree equation with the same roots.
and still have a second-degree equation with the same roots.

Current revision

A first-degree equation which has \displaystyle x=1+\sqrt{3} as a root is \displaystyle x-(1+\sqrt{3}\,)=0, which we can also write as \displaystyle x-1-\sqrt{3} = 0. In the same way, we have that \displaystyle x-(1-\sqrt{3}\,)=0, i.e., \displaystyle x-1+\sqrt{3}=0 is a first-degree equation that has \displaystyle x=1-\sqrt{3} as a root. If we multiply these two first-degree equations together, we get a second-degree equation with \displaystyle x=1+\sqrt{3} and \displaystyle x=1-\sqrt{3} as roots,

\displaystyle (x-1-\sqrt{3}\,)(x-1+\sqrt{3}\,) = 0\,\textrm{.}

The first factor become zero when \displaystyle x=1+\sqrt{3} and the second factor becomes zero when \displaystyle x=1-\sqrt{3}\,.

Nothing really prevents us from answering with \displaystyle (x-1-\sqrt{3}\,)(x-1+\sqrt{3}\,) = 0, but if we want to give the equation in standard form, we need to expand the left-hand side,

\displaystyle \begin{align}

(x-1-\sqrt{3}\,)(x-1+\sqrt{3}\,) &= x^{2} - x + \sqrt{3}x - x + 1 - \sqrt{3} - \sqrt{3}x + \sqrt{3} - (\sqrt{3}\,)^{2}\\[5pt] &= x^{2} + (-x+\sqrt{3}x-x-\sqrt{3}x) + (1-\sqrt{3}+\sqrt{3}-3)\\[5pt] &= x^{2}-2x-2 \end{align}

to get the equation \displaystyle x^{2}-2x-2=0\,.


Note: Exactly as in exercise a, we can multiply the equation by a non-zero constant a

\displaystyle ax^{2}-2ax-2a=0

and still have a second-degree equation with the same roots.

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