If-the-difference-of-two-roots-of-x-2-lx-m-0-is-1-then-prove-that-l-2-4m-2-1-2m-2-

Question Number 202459 by MATHEMATICSAM last updated on 27/Dec/23

If the difference of two roots of

x^{2} - l x + m = 0 is 1 then prove that

l^{2} + 4 m^{2} = {(1 + 2 m)}^{2} .

Answered by aleks041103 last updated on 27/Dec/23

x_{1, 2} = \frac{1}{2} (l \pm \sqrt{l^{2} - 4 m})

\Rightarrow∣ x_{1} - x_{2} ∣= \sqrt{l^{2} - 4 m} = 1

\Rightarrow l^{2} - 4 m = 1

\Rightarrow l^{2} = 1 + 4 m

\Rightarrow l^{2} + 4 m^{2} = 1 + 4 m + 4 m^{2} = 1 + 2 (1) (2 m) + {(2 m)}^{2}

\Rightarrow l^{2} + 4 m^{2} = {(1 + 2 m)}^{2}

Answered by Rasheed.Sindhi last updated on 27/Dec/23

r o o t s : α, α - 1 (s a y)

α + (α - 1) = l \land α (α - 1) = m

l = 2 α - 1 \land m = α (α - 1)

∙ l^{2} + 4 m^{2} = {(1 + 2 m)}^{2}

l h s : {(2 α - 1)}^{2} + 4 {(α (α - 1))}^{2}

= 4 α^{4} - 8 α^{3} + 8 α^{2} - 4 α + 1

r h s : {(1 + 2 α (α - 1))}^{2} = {(1 + 2 α^{2} - 2 α)}^{2}

= 4 α^{4} - 8 α^{3} + 8 α^{2} - 4 α + 1

∵ l h s = r h s

∴ p r o v e d

Answered by witcher3 last updated on 27/Dec/23

x_{2} - x_{1} = 1

\Rightarrow x_{2}^{2} + x_{1}^{2} - {2 x}_{1} x_{2} = 1

= x_{1}^{2} + x_{2}^{2} - 2 m = 1

x_{1}^{2} + x_{2}^{2} = {(x_{1} + x_{2})}^{2} - {2 x}_{1} x_{2} = l^{2} - 2 m

\Rightarrow l^{2} - 2 m - 2 m = 1

\Rightarrow l^{2} + {4 m}^{2} = 1 + 4 m + {4 m}^{2} = {(2 m + 1)}^{2}

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