contestada

Given that the roots of the equation a[tex]x^{2}[/tex]+b[tex]x[/tex]+c=0 are [tex]\beta[/tex] and n[tex]\beta[/tex], show that (n+1)²ac=nb²

Respuesta :

Answer:

Step-by-step explanation:

sum of roots β+nβ=-b/a

(n+1)β=-b/a

squaring

(n+1)²β²=b²/a²

product of roots β×nβ=c/a

nβ²=c/a

β²=c/na

∴(n+1)²c/na=b²/a²

multiply by na²

(n+1)²ac=nb²

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