Answer :
(16-x)(x+1)=15x
16x+16-x^2-x=15x
x^2-16=0
(x+4)(x-4)=0
x=+4,-4 Ans
16x+16-x^2-x=15x
x^2-16=0
(x+4)(x-4)=0
x=+4,-4 Ans
Answer:
The required roots of the polynomial is 4 and -4 respectively.
Step-by-step explanation:
Given That:-
[tex] : \implies \bf \: \frac{16}{x} - 1 = \frac{15}{ x + 1} \\ [/tex]
Here we have to find the roots for the given polynomial by using Quadratic Formula.
[tex] \implies \sf \: \frac{16}{x} - 1 = \frac{15}{ \\ x + 1} [/tex]
First make their denominators same.
[tex] \implies \sf \: \frac{16}{x} - \frac{1 \times x }{x} = \frac{15}{x + 1} \\ [/tex]
[tex] \implies \sf \: \frac{16 - x}{x} = \frac{15}{x + 1} \\ [/tex]
On doing cross multiplication on both sides we get,
[tex] \implies \sf \: (16 - x)(x + 1) = 15x[/tex]
[tex] \implies \sf \: 16x + 16 - {x}^{2} - x = 15x[/tex]
[tex] \implies \sf \: 15x + 16 - {x}^{2} = 15x[/tex]
[tex] \implies \sf \cancel{15x }+ 16 - {x}^{2} \cancel{- 15x }= 0[/tex]
[tex] \implies \sf \: 16 - {x}^{2} = 0[/tex]
Can also be written as,
[tex] \implies \sf \: {x}^{2} - 16 = 0[/tex]
Now, Formula For Finding Quadratic roots,
[tex]{ \bigstar{ \underline{ \boxed{ \bf{x = \frac{ - b \pm \sqrt{ {b}^{2} - 4ac } }{2a} }}}}} \bigstar[/tex]
Here consider a= 1 ; b = 0; c= -16.
[tex] \implies \sf \: x = \frac{0 \pm \sqrt{0 - 4(1)( - 16)} }{2(1)} \\ [/tex]
[tex] \implies \sf \: x = \frac{0 \pm \sqrt{0 + 64} }{2} \\ [/tex]
[tex] \implies \sf \: x = \frac{0 \pm \sqrt{64} }{2} \\ [/tex]
[tex] \implies \sf \: x = \frac{0 \pm 8}{2} \\ [/tex]
[tex] \implies \sf \: x = \frac{0 + 8}{2} \: \: ; \: \: \frac{0 - 8}{2} \\ [/tex]
[tex] \implies \sf \: x = \frac{8}{2} \: \: ; \: \: \frac{ - 8}{2} \\ [/tex]
[tex] \implies \sf \: x = \cancel{ \frac{8}{2}} \: \: ; \: \: \cancel{\frac{ - 8}{2} } \\ [/tex]
[tex] \implies \sf \: x = + 4 \: \: ; \: \: x = - 4[/tex]
Final Answer:-
[tex] {\therefore{ \boxed{ \bf{x = + 4 \: ;\: - 4}}}}[/tex]
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