Question

If zeroes of x^2-lx+m differ by 1 show that l^2=4m+1

Answer 1

$Given \: Quadratic \: polynomial \:x^{2}-lx+m \\and\: \alpha \:and \:\alpha - 1 \: are \: two\: zeroes$

$Compare \: the \: polynomial \:with \\ ax^{2}+bx+c, \: we\: get$

$a = 1 ,\: b = - l \:and \: c = m$

$i ) Sum \:of \:the \: zeroes = \frac{-b}{a}$

$\implies \alpha + \alpha - 1 = \frac{ -(-l) }{1 }$

$\implies 2 \alpha - 1 = l$

$\implies 2 \alpha = l + 1$

$\implies \alpha = \frac{l + 1 }{2} \: --(1)$

$ii ) Product\:of \:the \: zeroes = \frac{c}{a}$

$\implies \alpha ( \alpha - 1 )= \frac{ m }{1 }$

$\implies \Big( \frac{l + 1 }{2}\Big) [ \Big( \frac{l + 1 }{2}\Big) - 1 ]= m\: [ From \: (1) ]$

$\implies \Big( \frac{l + 1 }{2}\Big) [ \frac{l + 1 -2}{2} ] = m$

$\implies \Big( \frac{(l + 1) }{2}\Big) [ \frac{(l -1)}{2} ] = m$

$\implies \frac{\Big ( l^{2} - 1^{2}\Big)}{4} = m$

$\implies l^{2} - 1 = 4m$

$\implies l^{2} = 4m + 1$

$Hence\: proved$

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Answer 2

Answer:

\begin{gathered} Given \: Quadratic \: polynomial \:x^{2}-lx+m \\and\: \alpha \:and \:\alpha - 1 \: are \: two\: zeroes \end{gathered}GivenQuadraticpolynomialx2−lx+mandαandα−1aretwozeroes

\begin{gathered} Compare \: the \: polynomial \:with \\ ax^{2}+bx+c, \: we\: get \end{gathered}Comparethepolynomialwithax2+bx+c,weget

a = 1 ,\: b = - l \:and \: c = ma=1,b=−landc=m

i ) Sum \:of \:the \: zeroes = \frac{-b}{a}i)Sumofthezeroes=a−b

\implies \alpha + \alpha - 1 = \frac{ -(-l) }{1 }⟹α+α−1=1−(−l)

\implies 2 \alpha - 1 = l⟹2α−1=l

\implies 2 \alpha = l + 1⟹2α=l+1

\implies \alpha = \frac{l + 1 }{2} \: --(1)⟹α=2l+1−−(1)

ii ) Product\:of \:the \: zeroes = \frac{c}{a}ii)Productofthezeroes=ac

\implies \alpha ( \alpha - 1 )= \frac{ m }{1 }⟹α(α−1)=1m

\implies \Big( \frac{l + 1 }{2}\Big) [ \Big( \frac{l + 1 }{2}\Big) - 1 ]= m\: [ From \: (1) ]⟹(2l+1)[(2l+1)−1]=m[From(1)]

\implies \Big( \frac{l + 1 }{2}\Big) [ \frac{l + 1 -2}{2} ] = m⟹(2l+1)[2l+1−2]=m

\implies \Big( \frac{(l + 1) }{2}\Big) [ \frac{(l -1)}{2} ] = m⟹(2(l+1))[2(l−1)]=m

\implies \frac{\Big ( l^{2} - 1^{2}\Big)}{4} = m⟹4(l2−12)=m

\implies l^{2} - 1 = 4m⟹l2−1=4m

\implies l^{2} = 4m + 1⟹l2=4m+1

Hence\: provedHenceproved