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5x^2 + 19x - 4 = 0

 Nov 15, 2014

Best Answer 

 #2
avatar
+5

a=5 b=19 c=-4

we can use bhaskara:

Δ=b^2-4ac

Δ=19^2-4*5*(-4)

Δ=361+80=441

x'=$${\frac{\left({\mathtt{\,-\,}}{\mathtt{b}}{\mathtt{\,\small\textbf+\,}}{\sqrt{{\mathtt{441}}}}\right)}{\left({\mathtt{2}}{\mathtt{\,\times\,}}{\mathtt{a}}\right)}}$$ = $${\frac{\left({\mathtt{\,-\,}}{\mathtt{19}}{\mathtt{\,\small\textbf+\,}}{\mathtt{21}}\right)}{\left({\mathtt{2}}{\mathtt{\,\times\,}}{\mathtt{5}}\right)}}$$= 0.2

x''=$${\frac{\left({\mathtt{\,-\,}}{\mathtt{19}}{\mathtt{\,-\,}}{\mathtt{21}}\right)}{\left({\mathtt{10}}\right)}} = {\mathtt{\,-\,}}{\frac{{\mathtt{40}}}{{\mathtt{10}}}} = -{\mathtt{4}}$$

 

so the roots of the function are -4 and 0.2

 Nov 15, 2014
 #1
avatar+23254 
+5

5x² + 19x - 4  =  0

Let's see if we can factor this:  (5x - 1)(x + 4)  =  0

So, now either  5x - 1  =  0     or     x + 4  =  0

                            5x  =  1                  x  =  -4

                             x  =  1/5

 Nov 15, 2014
 #2
avatar
+5
Best Answer

a=5 b=19 c=-4

we can use bhaskara:

Δ=b^2-4ac

Δ=19^2-4*5*(-4)

Δ=361+80=441

x'=$${\frac{\left({\mathtt{\,-\,}}{\mathtt{b}}{\mathtt{\,\small\textbf+\,}}{\sqrt{{\mathtt{441}}}}\right)}{\left({\mathtt{2}}{\mathtt{\,\times\,}}{\mathtt{a}}\right)}}$$ = $${\frac{\left({\mathtt{\,-\,}}{\mathtt{19}}{\mathtt{\,\small\textbf+\,}}{\mathtt{21}}\right)}{\left({\mathtt{2}}{\mathtt{\,\times\,}}{\mathtt{5}}\right)}}$$= 0.2

x''=$${\frac{\left({\mathtt{\,-\,}}{\mathtt{19}}{\mathtt{\,-\,}}{\mathtt{21}}\right)}{\left({\mathtt{10}}\right)}} = {\mathtt{\,-\,}}{\frac{{\mathtt{40}}}{{\mathtt{10}}}} = -{\mathtt{4}}$$

 

so the roots of the function are -4 and 0.2

Guest Nov 15, 2014

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