Find k

Algebra 2

Find k

Postby nycmath » Fri Oct 02, 2026 6:42 pm

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nycmath
 
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Re: Find k

Postby Eigenvalue » Fri Oct 02, 2026 10:05 pm

[tex]\frac{ \sqrt{2}+ \sqrt{k} }{ \sqrt{2} }[/tex]=4

Multiply both sides by [tex]\sqrt{2}[/tex]
[tex]\sqrt{2}[/tex]+[tex]\sqrt{k}[/tex]=4[tex]\sqrt{2}[/tex]
[tex]\sqrt{k}[/tex]=3[tex]\sqrt{2}[/tex]
k=(3[tex]\sqrt{2}[/tex])²=18

Find the polynomial:
The quadratic formula is x=[tex]\frac{-b± \sqrt{b²-4ac} }{2a}[/tex]

The expression pertains to the root [tex]\frac{-b+\sqrt{b²-4ac} }{2a}[/tex]

-b=[tex]\sqrt{2}[/tex]
b=-[tex]\sqrt{2}[/tex]

2a=[tex]\sqrt{2}[/tex]

a=[tex]\frac{ \sqrt{2} }{2}[/tex]

b²-4ac=18
(-[tex]\sqrt{2}[/tex])²-4([tex]\frac{ \sqrt{2} }{2}[/tex])c=18
2-2c[tex]\sqrt{2}[/tex]=18

-2c[tex]\sqrt{2}[/tex]=16

c[tex]\sqrt{2}[/tex]=-8

c=[tex]\frac{-8 \sqrt{2} }{2}[/tex]=-4[tex]\sqrt{2}[/tex]

The polynomial is [tex]\frac{ \sqrt{2} }{2}[/tex]x²-[tex]\sqrt{2}[/tex]x-4[tex]\sqrt{2}[/tex]

Eigenvalue
 
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