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Question Number 55014 by Rio Mike last updated on 16/Feb/19

what is the value of t that makes   x^2 +10x+t a perfect square?

$${what}\:{is}\:{the}\:{value}\:{of}\:{t}\:{that}\:{makes}\: \\ $$$${x}^{\mathrm{2}} +\mathrm{10}{x}+{t}\:{a}\:{perfect}\:{square}? \\ $$$$ \\ $$$$ \\ $$$$ \\ $$

Answered by peter frank last updated on 16/Feb/19

t=25

$${t}=\mathrm{25} \\ $$

Answered by Joel578 last updated on 16/Feb/19

x^2  + 10x + t ≡ (x + (√t))^2   x^2  + 10x + t ≡ x^2  + 2(√t)x + t  10 = 2(√t) → t = 25

$${x}^{\mathrm{2}} \:+\:\mathrm{10}{x}\:+\:{t}\:\equiv\:\left({x}\:+\:\sqrt{{t}}\right)^{\mathrm{2}} \\ $$$${x}^{\mathrm{2}} \:+\:\mathrm{10}{x}\:+\:{t}\:\equiv\:{x}^{\mathrm{2}} \:+\:\mathrm{2}\sqrt{{t}}{x}\:+\:{t} \\ $$$$\mathrm{10}\:=\:\mathrm{2}\sqrt{{t}}\:\rightarrow\:{t}\:=\:\mathrm{25} \\ $$

Answered by tanmay.chaudhury50@gmail.com last updated on 16/Feb/19

x^2 +2×x×5+5^2 +t−25  (x+5)^2 +(t−25)  to make perfect square (t−25)=0  so t=25

$${x}^{\mathrm{2}} +\mathrm{2}×{x}×\mathrm{5}+\mathrm{5}^{\mathrm{2}} +{t}−\mathrm{25} \\ $$$$\left({x}+\mathrm{5}\right)^{\mathrm{2}} +\left({t}−\mathrm{25}\right) \\ $$$${to}\:{make}\:{perfect}\:{square}\:\left({t}−\mathrm{25}\right)=\mathrm{0} \\ $$$${so}\:{t}=\mathrm{25} \\ $$

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