The-number-of-real-roots-of-the-quadratic-equation-x-4-2-x-5-2-x-6-2-0-is-

Question Number 98056 by PengagumRahasiamu last updated on 11/Jun/20

The number of real roots of the quadratic

equation {(x - 4)}^{2} + {(x - 5)}^{2} + {(x - 6)}^{2} = 0 is

Commented by som(math1967) last updated on 11/Jun/20

4, 5, 6

sum of 3 squares are 0

∴ each of them 0

∴ {(x - 4)}^{2} = 0 \Rightarrow x = 4

{(x - 5)}^{2} = 0 \Rightarrow x = 5

{(x - 6)}^{2} = 0 \Rightarrow x = 6

Commented by mr W last updated on 11/Jun/20

t h e r e i s n o r e a l r o o t!

l e t t = x - 5

{(x - 4)}^{2} + {(x - 5)}^{2} + {(x - 6)}^{2}

= {(t + 1)}^{2} + t^{2} + {(t - 1)}^{2}

= 3 t^{2} + 2

⩾ 2

i . e . {(x - 4)}^{2} + {(x - 5)}^{2} + {(x - 6)}^{2} = 0 h a s

n o r e a l r o o t!

Commented by bemath last updated on 11/Jun/20

let x - 6 = t - 1, x - 5 = t, x - 4 = t + 1

\Leftrightarrow t^{2} + {(t + 1)}^{2} + {(t - 1)}^{2} = 0

t^{2} + t^{2} + t^{2} + 2 = 0

{3 t}^{2} = - 2 \Leftrightarrow t \notin R, since t^{2} ⩾ 0

for t \in R

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