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Let f(x) be a quadratic polynomial with integer coefficients such that f(0) and f(1) are odd integers. Prove that the equation f(x) = 0 does not have an integer solution.

$$\mathrm{Let}\:{f}\left({x}\right)\:\mathrm{be}\:\mathrm{a}\:\mathrm{quadratic}\:\mathrm{polynomial} \\ $$$$\mathrm{with}\:\mathrm{integer}\:\mathrm{coefficients}\:\mathrm{such}\:\mathrm{that}\:{f}\left(\mathrm{0}\right) \\ $$$$\mathrm{and}\:{f}\left(\mathrm{1}\right)\:\mathrm{are}\:\mathrm{odd}\:\mathrm{integers}.\:\mathrm{Prove}\:\mathrm{that} \\ $$$$\mathrm{the}\:\mathrm{equation}\:{f}\left({x}\right)\:=\:\mathrm{0}\:\mathrm{does}\:\mathrm{not}\:\mathrm{have}\:\mathrm{an} \\ $$$$\mathrm{integer}\:\mathrm{solution}. \\ $$

Question Number 19215 Answers: 1 Comments: 0

STATEMENT-1 : For every natural number n ≥ 2, (1/(√1)) + (1/(√2)) + ..... (1/(√n)) > (√n) and STATEMENT-2 : For every natural number n ≥ 2, (√(n(n + 1))) < n + 1

$$\mathrm{STATEMENT}-\mathrm{1}\::\:\mathrm{For}\:\mathrm{every}\:\mathrm{natural} \\ $$$$\mathrm{number}\:{n}\:\geqslant\:\mathrm{2},\:\frac{\mathrm{1}}{\sqrt{\mathrm{1}}}\:+\:\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}\:+\:.....\:\frac{\mathrm{1}}{\sqrt{{n}}}\:>\:\sqrt{{n}} \\ $$$$\boldsymbol{\mathrm{and}} \\ $$$$\mathrm{STATEMENT}-\mathrm{2}\::\:\mathrm{For}\:\mathrm{every}\:\mathrm{natural} \\ $$$$\mathrm{number}\:{n}\:\geqslant\:\mathrm{2},\:\sqrt{{n}\left({n}\:+\:\mathrm{1}\right)}\:<\:{n}\:+\:\mathrm{1} \\ $$

Question Number 19204 Answers: 0 Comments: 1

If L= [((1 0 0)),((3 1 0)),((2 4 1)) ]and B= [(3),(2),(1) ] x_1 =−2 ; x_2 =1 ; x_3 =5 find U (numerical analysis)

$$\mathrm{If}\:\mathrm{L}=\begin{bmatrix}{\mathrm{1}\:\:\mathrm{0}\:\:\mathrm{0}}\\{\mathrm{3}\:\:\mathrm{1}\:\:\mathrm{0}}\\{\mathrm{2}\:\:\mathrm{4}\:\:\mathrm{1}}\end{bmatrix}\mathrm{and}\:\mathrm{B}=\begin{bmatrix}{\mathrm{3}}\\{\mathrm{2}}\\{\mathrm{1}}\end{bmatrix} \\ $$$$\mathrm{x}_{\mathrm{1}} =−\mathrm{2}\:;\:\mathrm{x}_{\mathrm{2}} =\mathrm{1}\:;\:\mathrm{x}_{\mathrm{3}} =\mathrm{5} \\ $$$$\mathrm{find}\:\mathrm{U} \\ $$$$\left(\mathrm{numerical}\:\mathrm{analysis}\right) \\ $$$$ \\ $$

Question Number 19198 Answers: 1 Comments: 7

Find all integer solutions of the system: 35x + 63y + 45z = 1, ∣x∣ < 9, ∣y∣ < 5, ∣z∣ < 7.

$$\mathrm{Find}\:\mathrm{all}\:\mathrm{integer}\:\mathrm{solutions}\:\mathrm{of}\:\mathrm{the}\:\mathrm{system}: \\ $$$$\mathrm{35}{x}\:+\:\mathrm{63}{y}\:+\:\mathrm{45}{z}\:=\:\mathrm{1},\:\mid{x}\mid\:<\:\mathrm{9},\:\mid{y}\mid\:<\:\mathrm{5}, \\ $$$$\mid{z}\mid\:<\:\mathrm{7}. \\ $$

Question Number 19171 Answers: 1 Comments: 1

log_(√2) (√(2(√(2(√(2(√(2 ))))))))

$${log}_{\sqrt{\mathrm{2}}} \:\sqrt{\mathrm{2}\sqrt{\mathrm{2}\sqrt{\mathrm{2}\sqrt{\mathrm{2}\:\:\:\:}}}} \\ $$

Question Number 19135 Answers: 1 Comments: 0

solve for x: 2^(∣x+2∣) −∣2^(x+1) −1∣=2^(x+1) +1

$${solve}\:{for}\:{x}: \\ $$$$\mathrm{2}^{\mid{x}+\mathrm{2}\mid} −\mid\mathrm{2}^{{x}+\mathrm{1}} −\mathrm{1}\mid=\mathrm{2}^{{x}+\mathrm{1}} +\mathrm{1} \\ $$

Question Number 19123 Answers: 1 Comments: 0

{ ((xf(x)−g(x)+h(x)=2x+1)),((f(x)−(2x−2)g(x)−3h(x)=x)),((ln (x)f(x)−(x−3)h(x)=1)) :} Find f(x),g(x),h(x)

Question Number 19121 Answers: 0 Comments: 0

Question Number 19101 Answers: 0 Comments: 3

A polynomial f(x) with rational coefficients leaves remainder 15, when divided by x − 3 and remainder 2x + 1, when divided by (x − 1)^2 . Find the remainder when f(x) is divided by (x − 3)(x − 1)^2 .

$$\mathrm{A}\:\mathrm{polynomial}\:{f}\left({x}\right)\:\mathrm{with}\:\mathrm{rational} \\ $$$$\mathrm{coefficients}\:\mathrm{leaves}\:\mathrm{remainder}\:\mathrm{15},\:\mathrm{when} \\ $$$$\mathrm{divided}\:\mathrm{by}\:{x}\:−\:\mathrm{3}\:\mathrm{and}\:\mathrm{remainder}\:\mathrm{2}{x}\:+\:\mathrm{1}, \\ $$$$\mathrm{when}\:\mathrm{divided}\:\mathrm{by}\:\left({x}\:−\:\mathrm{1}\right)^{\mathrm{2}} .\:\mathrm{Find}\:\mathrm{the} \\ $$$$\mathrm{remainder}\:\mathrm{when}\:{f}\left({x}\right)\:\mathrm{is}\:\mathrm{divided}\:\mathrm{by} \\ $$$$\left({x}\:−\:\mathrm{3}\right)\left({x}\:−\:\mathrm{1}\right)^{\mathrm{2}} . \\ $$

Question Number 19095 Answers: 0 Comments: 3