Mathematics: Which equation can be used to solve for b? 5 cm 10 cm tan(309) = 5 tan(309) = tan(309) = 10 tan(30)=10

Step-by-step explanation:We have the following quadratic equation:Which is in the form: And can be solved with the quadratic formula: Where , , Knowing this, let s begin:Factoring we have . hence:Solving for both options:...

Which equation can be used to solve for b? 5

Ответ:

crazymadhatter0

01.01.2020

Mathematics

$x=\frac{(1-k^{2}) \pm (1+k^{2})}{2k}$

Step-by-step explanation:

We have the following quadratic equation:

$x^{2} + (k-\frac{1}{k})x- 1=0$

Which is in the form: $ax^{2} + bx + c=0$

And can be solved with the quadratic formula: $x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$

Where a=1 , $b=k-\frac{1}{k}$ , c=1

Knowing this, let's begin:

$x=\frac{-(k-\frac{1}{k})\pm\sqrt{(k-\frac{1}{k})^{2}-4(1)(-1)}}{2(1)}$

$x=\frac{-k+\frac{1}{k}\pm\sqrt{k^{2}-2k\frac{1}{k}+\frac{1}{k^{2}} +4}}{2}$

$x=\frac{\frac{1-k^{2}}{k}\pm\sqrt{k^{2}+\frac{1}{k^{2}} +2}}{2}$

$x=\frac{1-k^{2}}{2k} \pm \frac{\sqrt{\frac{k^{4}+1+2k^{2}}{k^{2}}}}{2}$

$x=\frac{1-k^{2}}{2k} \pm \frac{\sqrt{k^{4}+2k^{2} +1}}{2k}$

Factoring $k^{4}+2k^{2} +1$ we have $(k^{2} +1 )^{2}$ . hence:

$x=\frac{1-k^{2}}{2k} \pm \frac{\sqrt{(k^{2}+1)^{2}}}{2k}$

$x=\frac{(1-k^{2}) \pm (1+k^{2})}{2k}$

Solving for both options:

$x_{1}=\frac{(1-k^{2}) + (1+k^{2})}{2k}=\frac{1}{k}$

$x_{2}=\frac{(1-k^{2}) - (1+k^{2})}{2k}=-k$

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Which equation can be used to solve for b? 5 cm 10 cm tan(309) = 5 tan(309) = tan(309) = 10 tan(30)=10

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