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AlgebraQuestion and Answers: Page 347

Question Number 20551 Answers: 1 Comments: 8

Find the minimum value of ∣a + bω + cω^2 ∣, where a, b and c are all not equal integers and ω(≠1) is a cube root of unity.

$${Find}\:{the}\:{minimum}\:{value}\:{of} \\ $$$$\mid{a}\:+\:{b}\omega\:+\:{c}\omega^{\mathrm{2}} \mid,\:{where}\:{a},\:{b}\:{and}\:{c}\:{are}\:{all} \\ $$$${not}\:{equal}\:{integers}\:{and}\:\omega\left(\neq\mathrm{1}\right)\:{is}\:{a}\:{cube} \\ $$$${root}\:{of}\:{unity}. \\ $$

Question Number 20488 Answers: 1 Comments: 1

Question Number 20430 Answers: 2 Comments: 1

Let a, b and c be such that a + b + c = 0 and P = (a^2 /(2a^2 + bc)) + (b^2 /(2b^2 + ca)) + (c^2 /(2c^2 + ab)) is defined. What is the value of P?

$$\mathrm{Let}\:{a},\:{b}\:\mathrm{and}\:{c}\:\mathrm{be}\:\mathrm{such}\:\mathrm{that}\:{a}\:+\:{b}\:+\:{c}\:=\:\mathrm{0} \\ $$$$\mathrm{and} \\ $$$${P}\:=\:\frac{{a}^{\mathrm{2}} }{\mathrm{2}{a}^{\mathrm{2}} \:+\:{bc}}\:+\:\frac{{b}^{\mathrm{2}} }{\mathrm{2}{b}^{\mathrm{2}} \:+\:{ca}}\:+\:\frac{{c}^{\mathrm{2}} }{\mathrm{2}{c}^{\mathrm{2}} \:+\:{ab}} \\ $$$$\mathrm{is}\:\mathrm{defined}.\:\mathrm{What}\:\mathrm{is}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:{P}? \\ $$

Question Number 20372 Answers: 2 Comments: 0

The two roots of an equation x^3 − 9x^2 + 14x + 24 = 0 are in the ratio 3 : 2. Find the roots.

$$\mathrm{The}\:\mathrm{two}\:\mathrm{roots}\:\mathrm{of}\:\mathrm{an}\:\mathrm{equation}\:{x}^{\mathrm{3}} \:−\:\mathrm{9}{x}^{\mathrm{2}} \\ $$$$+\:\mathrm{14}{x}\:+\:\mathrm{24}\:=\:\mathrm{0}\:\mathrm{are}\:\mathrm{in}\:\mathrm{the}\:\mathrm{ratio}\:\mathrm{3}\::\:\mathrm{2}. \\ $$$$\mathrm{Find}\:\mathrm{the}\:\mathrm{roots}. \\ $$

Question Number 20367 Answers: 1 Comments: 0

Let f(x) = x^3 + 3x^2 + 9x + 6sinx, then find the number of real roots of the equation (1/(x − f(1))) + (2/(x − f(2))) + (3/(x − f(3))) = 0.

$$\mathrm{Let}\:{f}\left({x}\right)\:=\:{x}^{\mathrm{3}} \:+\:\mathrm{3}{x}^{\mathrm{2}} \:+\:\mathrm{9}{x}\:+\:\mathrm{6sin}{x},\:\mathrm{then} \\ $$$$\mathrm{find}\:\mathrm{the}\:\mathrm{number}\:\mathrm{of}\:\mathrm{real}\:\mathrm{roots}\:\mathrm{of}\:\mathrm{the} \\ $$$$\mathrm{equation} \\ $$$$\frac{\mathrm{1}}{{x}\:−\:{f}\left(\mathrm{1}\right)}\:+\:\frac{\mathrm{2}}{{x}\:−\:{f}\left(\mathrm{2}\right)}\:+\:\frac{\mathrm{3}}{{x}\:−\:{f}\left(\mathrm{3}\right)}\:=\:\mathrm{0}. \\ $$

Question Number 20368 Answers: 0 Comments: 2

If α is a real root of 2x^3 − 3x^2 + 6x + 6 = 0, then find [α] where [∙] denotes the greatest integer function.

$$\mathrm{If}\:\alpha\:\mathrm{is}\:\mathrm{a}\:\mathrm{real}\:\mathrm{root}\:\mathrm{of}\:\mathrm{2}{x}^{\mathrm{3}} \:−\:\mathrm{3}{x}^{\mathrm{2}} \:+\:\mathrm{6}{x}\:+\:\mathrm{6}\:=\:\mathrm{0}, \\ $$$$\mathrm{then}\:\mathrm{find}\:\left[\alpha\right]\:\mathrm{where}\:\left[\centerdot\right]\:\mathrm{denotes}\:\mathrm{the} \\ $$$$\mathrm{greatest}\:\mathrm{integer}\:\mathrm{function}. \\ $$

Question Number 20308 Answers: 1 Comments: 0

For what value of k, (x + y + z)^2 + k(x^2 + y^2 + z^2 ) can be resolved into linear rational factors?

$$\mathrm{For}\:\mathrm{what}\:\mathrm{value}\:\mathrm{of}\:{k},\:\left({x}\:+\:{y}\:+\:{z}\right)^{\mathrm{2}} \:+ \\ $$$${k}\left({x}^{\mathrm{2}} \:+\:{y}^{\mathrm{2}} \:+\:{z}^{\mathrm{2}} \right)\:\mathrm{can}\:\mathrm{be}\:\mathrm{resolved}\:\mathrm{into} \\ $$$$\mathrm{linear}\:\mathrm{rational}\:\mathrm{factors}? \\ $$

Question Number 20307 Answers: 2 Comments: 0

Show that a(b − c)x^2 + b(c − a)xy + c(a − b)y^2 will be a perfect square if a, b, c are in H.P.

$$\mathrm{Show}\:\mathrm{that}\:{a}\left({b}\:−\:{c}\right){x}^{\mathrm{2}} \:+\:{b}\left({c}\:−\:{a}\right){xy}\:+ \\ $$$${c}\left({a}\:−\:{b}\right){y}^{\mathrm{2}} \:\mathrm{will}\:\mathrm{be}\:\mathrm{a}\:\mathrm{perfect}\:\mathrm{square}\:\mathrm{if}\:{a}, \\ $$$${b},\:{c}\:\mathrm{are}\:\mathrm{in}\:\mathrm{H}.\mathrm{P}. \\ $$

Question Number 20298 Answers: 1 Comments: 0

Question Number 20297 Answers: 1 Comments: 0

Prove that the expression ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0 can be resolved into two linear rational factors if Δ = abc + 2fgh − af^2 − bg^2 − ch^2 = 0

$$\mathrm{Prove}\:\mathrm{that}\:\mathrm{the}\:\mathrm{expression}\:{ax}^{\mathrm{2}} \:+\:\mathrm{2}{hxy} \\ $$$$+\:{by}^{\mathrm{2}} \:+\:\mathrm{2}{gx}\:+\:\mathrm{2}{fy}\:+\:{c}\:=\:\mathrm{0}\:\mathrm{can}\:\mathrm{be} \\ $$$$\mathrm{resolved}\:\mathrm{into}\:\mathrm{two}\:\mathrm{linear}\:\mathrm{rational}\:\mathrm{factors} \\ $$$$\mathrm{if}\:\Delta\:=\:{abc}\:+\:\mathrm{2}{fgh}\:−\:{af}^{\mathrm{2}} \:−\:{bg}^{\mathrm{2}} \:−\:{ch}^{\mathrm{2}} \:=\:\mathrm{0} \\ $$

Question Number 20296 Answers: 2 Comments: 0

If (m_r , (1/m_r )) ; r = 1, 2, 3, 4 be four pairs of values of x and y satisfy the equation x^2 + y^2 + 2gx + 2fy + c = 0, then prove that m_1 .m_2 .m_3 .m_4 = 1.

$$\mathrm{If}\:\left({m}_{{r}} \:,\:\frac{\mathrm{1}}{{m}_{{r}} }\right)\:;\:{r}\:=\:\mathrm{1},\:\mathrm{2},\:\mathrm{3},\:\mathrm{4}\:\mathrm{be}\:\mathrm{four}\:\mathrm{pairs} \\ $$$$\mathrm{of}\:\mathrm{values}\:\mathrm{of}\:{x}\:\mathrm{and}\:{y}\:\mathrm{satisfy}\:\mathrm{the}\:\mathrm{equation} \\ $$$${x}^{\mathrm{2}} \:+\:{y}^{\mathrm{2}} \:+\:\mathrm{2}{gx}\:+\:\mathrm{2}{fy}\:+\:{c}\:=\:\mathrm{0},\:\mathrm{then}\:\mathrm{prove} \\ $$$$\mathrm{that}\:{m}_{\mathrm{1}} .{m}_{\mathrm{2}} .{m}_{\mathrm{3}} .{m}_{\mathrm{4}} \:=\:\mathrm{1}. \\ $$

Question Number 20259 Answers: 1 Comments: 0

Find the number of real roots of the equation f(x) = x^3 + 2x^2 + 2x + 1 = 0

$$\mathrm{Find}\:\mathrm{the}\:\mathrm{number}\:\mathrm{of}\:\mathrm{real}\:\mathrm{roots}\:\mathrm{of}\:\mathrm{the} \\ $$$$\mathrm{equation}\:{f}\left({x}\right)\:=\:{x}^{\mathrm{3}} \:+\:\mathrm{2}{x}^{\mathrm{2}} \:+\:\mathrm{2}{x}\:+\:\mathrm{1}\:=\:\mathrm{0} \\ $$

Question Number 20118 Answers: 1 Comments: 0

The quadratic equations x^2 − 6x + a = 0 and x^2 − cx + 6 = 0 have one root in common. The other roots of the first and second equations are integers in the ratio 4 : 3. Then, find the common root.

$$\mathrm{The}\:\mathrm{quadratic}\:\mathrm{equations}\:{x}^{\mathrm{2}} \:−\:\mathrm{6}{x}\:+\:{a}\:=\:\mathrm{0} \\ $$$$\mathrm{and}\:{x}^{\mathrm{2}} \:−\:{cx}\:+\:\mathrm{6}\:=\:\mathrm{0}\:\mathrm{have}\:\mathrm{one}\:\mathrm{root}\:\mathrm{in} \\ $$$$\mathrm{common}.\:\mathrm{The}\:\mathrm{other}\:\mathrm{roots}\:\mathrm{of}\:\mathrm{the}\:\mathrm{first} \\ $$$$\mathrm{and}\:\mathrm{second}\:\mathrm{equations}\:\mathrm{are}\:\mathrm{integers}\:\mathrm{in} \\ $$$$\mathrm{the}\:\mathrm{ratio}\:\mathrm{4}\::\:\mathrm{3}.\:\mathrm{Then},\:\mathrm{find}\:\mathrm{the}\:\mathrm{common} \\ $$$$\mathrm{root}. \\ $$

Question Number 20116 Answers: 1 Comments: 0

If a and b (≠ 0) are the roots of the equation x^2 + ax + b = 0, then find the least value of x^2 + ax + b (x ∈ R).

$$\mathrm{If}\:{a}\:\mathrm{and}\:{b}\:\left(\neq\:\mathrm{0}\right)\:\mathrm{are}\:\mathrm{the}\:\mathrm{roots}\:\mathrm{of}\:\mathrm{the} \\ $$$$\mathrm{equation}\:{x}^{\mathrm{2}} \:+\:{ax}\:+\:{b}\:=\:\mathrm{0},\:\mathrm{then}\:\mathrm{find}\:\mathrm{the} \\ $$$$\mathrm{least}\:\mathrm{value}\:\mathrm{of}\:{x}^{\mathrm{2}} \:+\:{ax}\:+\:{b}\:\left({x}\:\in\:{R}\right). \\ $$

Question Number 20115 Answers: 1 Comments: 0

The value of a for which the equation (1 − a^2 )x^2 + 2ax − 1 = 0 has roots belonging to (0, 1) is

$$\mathrm{The}\:\mathrm{value}\:\mathrm{of}\:{a}\:\mathrm{for}\:\mathrm{which}\:\mathrm{the}\:\mathrm{equation} \\ $$$$\left(\mathrm{1}\:−\:{a}^{\mathrm{2}} \right){x}^{\mathrm{2}} \:+\:\mathrm{2}{ax}\:−\:\mathrm{1}\:=\:\mathrm{0}\:\mathrm{has}\:\mathrm{roots} \\ $$$$\mathrm{belonging}\:\mathrm{to}\:\left(\mathrm{0},\:\mathrm{1}\right)\:\mathrm{is} \\ $$

Question Number 20054 Answers: 1 Comments: 0

If α and β (α < β) are the roots of the equation x^2 + bx + c = 0, where c < 0 < b, then (1) 0 < α < β (2) α < 0 < β < ∣α∣ (3) α < β < 0 (4) α < 0 < ∣α∣ < β

$$\mathrm{If}\:\alpha\:\mathrm{and}\:\beta\:\left(\alpha\:<\:\beta\right)\:\mathrm{are}\:\mathrm{the}\:\mathrm{roots}\:\mathrm{of}\:\mathrm{the} \\ $$$$\mathrm{equation}\:{x}^{\mathrm{2}} \:+\:{bx}\:+\:{c}\:=\:\mathrm{0},\:\mathrm{where} \\ $$$${c}\:<\:\mathrm{0}\:<\:{b},\:\mathrm{then} \\ $$$$\left(\mathrm{1}\right)\:\mathrm{0}\:<\:\alpha\:<\:\beta \\ $$$$\left(\mathrm{2}\right)\:\alpha\:<\:\mathrm{0}\:<\:\beta\:<\:\mid\alpha\mid \\ $$$$\left(\mathrm{3}\right)\:\alpha\:<\:\beta\:<\:\mathrm{0} \\ $$$$\left(\mathrm{4}\right)\:\alpha\:<\:\mathrm{0}\:<\:\mid\alpha\mid\:<\:\beta \\ $$

Question Number 20053 Answers: 1 Comments: 0

If (4a + c)^2 ≤ 4b^2 then one root of ax^2 + bx + c = 0 lies in (1) (−2, 2) (2) (−1, 1) (3) (−∞, −2) (4) (2, ∞)

$$\mathrm{If}\:\left(\mathrm{4}{a}\:+\:{c}\right)^{\mathrm{2}} \:\leqslant\:\mathrm{4}{b}^{\mathrm{2}} \:\mathrm{then}\:\mathrm{one}\:\mathrm{root}\:\mathrm{of} \\ $$$${ax}^{\mathrm{2}} \:+\:{bx}\:+\:{c}\:=\:\mathrm{0}\:\mathrm{lies}\:\mathrm{in} \\ $$$$\left(\mathrm{1}\right)\:\left(−\mathrm{2},\:\mathrm{2}\right) \\ $$$$\left(\mathrm{2}\right)\:\left(−\mathrm{1},\:\mathrm{1}\right) \\ $$$$\left(\mathrm{3}\right)\:\left(−\infty,\:−\mathrm{2}\right) \\ $$$$\left(\mathrm{4}\right)\:\left(\mathrm{2},\:\infty\right) \\ $$

Question Number 20052 Answers: 1 Comments: 0

If the roots α and β of the equation ax^2 + bx + c = 0 are real and of opposite sign then the roots of the equation α(x − β)^2 + β(x − α)^2 is/are (1) Positive (2) Negative (3) Real and opposite sign (4) Imaginary

$$\mathrm{If}\:\mathrm{the}\:\mathrm{roots}\:\alpha\:\mathrm{and}\:\beta\:\mathrm{of}\:\mathrm{the}\:\mathrm{equation} \\ $$$${ax}^{\mathrm{2}} \:+\:{bx}\:+\:{c}\:=\:\mathrm{0}\:\mathrm{are}\:\mathrm{real}\:\mathrm{and}\:\mathrm{of}\:\mathrm{opposite} \\ $$$$\mathrm{sign}\:\mathrm{then}\:\mathrm{the}\:\mathrm{roots}\:\mathrm{of}\:\mathrm{the}\:\mathrm{equation} \\ $$$$\alpha\left({x}\:−\:\beta\right)^{\mathrm{2}} \:+\:\beta\left({x}\:−\:\alpha\right)^{\mathrm{2}} \:\mathrm{is}/\mathrm{are} \\ $$$$\left(\mathrm{1}\right)\:\mathrm{Positive} \\ $$$$\left(\mathrm{2}\right)\:\mathrm{Negative} \\ $$$$\left(\mathrm{3}\right)\:\mathrm{Real}\:\mathrm{and}\:\mathrm{opposite}\:\mathrm{sign} \\ $$$$\left(\mathrm{4}\right)\:\mathrm{Imaginary} \\ $$

Question Number 20047 Answers: 1 Comments: 0

Solve for x: ((√(x + 1))/x) + (√(x/(x + 1))) = ((13)/6)

$$\mathrm{Solve}\:\mathrm{for}\:\mathrm{x}: \\ $$$$\frac{\sqrt{\mathrm{x}\:+\:\mathrm{1}}}{\mathrm{x}}\:+\:\sqrt{\frac{\mathrm{x}}{\mathrm{x}\:+\:\mathrm{1}}}\:=\:\frac{\mathrm{13}}{\mathrm{6}} \\ $$

Question Number 20051 Answers: 1 Comments: 0

If x ∈ R then ((x^2 + 2x + a)/(x^2 + 4x + 3a)) can take all real values if (1) a ∈ (0, 2) (2) a ∈ [0, 1] (3) a ∈ [−1, 1] (4) None of these

$$\mathrm{If}\:{x}\:\in\:{R}\:\mathrm{then}\:\frac{{x}^{\mathrm{2}} \:+\:\mathrm{2}{x}\:+\:{a}}{{x}^{\mathrm{2}} \:+\:\mathrm{4}{x}\:+\:\mathrm{3}{a}}\:\mathrm{can}\:\mathrm{take}\:\mathrm{all} \\ $$$$\mathrm{real}\:\mathrm{values}\:\mathrm{if} \\ $$$$\left(\mathrm{1}\right)\:{a}\:\in\:\left(\mathrm{0},\:\mathrm{2}\right) \\ $$$$\left(\mathrm{2}\right)\:{a}\:\in\:\left[\mathrm{0},\:\mathrm{1}\right] \\ $$$$\left(\mathrm{3}\right)\:{a}\:\in\:\left[−\mathrm{1},\:\mathrm{1}\right] \\ $$$$\left(\mathrm{4}\right)\:\mathrm{None}\:\mathrm{of}\:\mathrm{these} \\ $$

Question Number 20013 Answers: 0 Comments: 4

Show that the equation (1/(x − a)) + (1/(x − b)) + (1/(x − c)) = 0 can have a pair of equal roots if a = b = c.

$$\mathrm{Show}\:\mathrm{that}\:\mathrm{the}\:\mathrm{equation}\:\frac{\mathrm{1}}{{x}\:−\:{a}}\:+\:\frac{\mathrm{1}}{{x}\:−\:{b}} \\ $$$$+\:\frac{\mathrm{1}}{{x}\:−\:{c}}\:=\:\mathrm{0}\:\mathrm{can}\:\mathrm{have}\:\mathrm{a}\:\mathrm{pair}\:\mathrm{of}\:\mathrm{equal} \\ $$$$\mathrm{roots}\:\mathrm{if}\:{a}\:=\:{b}\:=\:{c}. \\ $$

Question Number 20001 Answers: 1 Comments: 0

The number of the roots of the quadratic equation 8sec^2 θ − 6secθ + 1 = 0 is

$$\mathrm{The}\:\mathrm{number}\:\mathrm{of}\:\mathrm{the}\:\mathrm{roots}\:\mathrm{of}\:\mathrm{the}\:\mathrm{quadratic} \\ $$$$\mathrm{equation}\:\mathrm{8sec}^{\mathrm{2}} \theta\:−\:\mathrm{6sec}\theta\:+\:\mathrm{1}\:=\:\mathrm{0}\:\mathrm{is} \\ $$

Question Number 19945 Answers: 1 Comments: 0

If α and β are the roots of equation x^2 + px + q = 0 and α^2 , β^2 are roots of the equation x^2 − rx + s = 0, show that the equation x^2 − 4qx + 2q^2 − r = 0 has real roots.

$$\mathrm{If}\:\alpha\:\mathrm{and}\:\beta\:\mathrm{are}\:\mathrm{the}\:\mathrm{roots}\:\mathrm{of}\:\mathrm{equation} \\ $$$${x}^{\mathrm{2}} \:+\:{px}\:+\:{q}\:=\:\mathrm{0}\:\mathrm{and}\:\alpha^{\mathrm{2}} ,\:\beta^{\mathrm{2}} \:\mathrm{are}\:\mathrm{roots}\:\mathrm{of} \\ $$$$\mathrm{the}\:\mathrm{equation}\:{x}^{\mathrm{2}} \:−\:{rx}\:+\:{s}\:=\:\mathrm{0},\:\mathrm{show} \\ $$$$\mathrm{that}\:\mathrm{the}\:\mathrm{equation}\:{x}^{\mathrm{2}} \:−\:\mathrm{4}{qx}\:+\:\mathrm{2}{q}^{\mathrm{2}} \:−\:{r}\:=\:\mathrm{0} \\ $$$$\mathrm{has}\:\mathrm{real}\:\mathrm{roots}. \\ $$

Question Number 19936 Answers: 1 Comments: 1

Find the pricipal value of (1−i)^(1+i) .

$${Find}\:{the}\:{pricipal}\:{value}\:{of}\: \\ $$$$\:\:\:\:\:\left(\mathrm{1}−{i}\right)^{\mathrm{1}+{i}} \:. \\ $$

Question Number 19914 Answers: 0 Comments: 0

Question Number 19900 Answers: 2 Comments: 0

Prove that this is an identity in x: (((x−a)(x−b))/((c−a)(c−b)))+(((x−b)(x−c))/((a−b)(a−c)))+(((x−c)(x−a))/((b−c)(b−a)))=1

$$\mathrm{Prove}\:\mathrm{that}\:\mathrm{this}\:\mathrm{is}\:\mathrm{an}\:\mathrm{identity}\:\mathrm{in}\:{x}: \\ $$$$\frac{\left({x}−{a}\right)\left({x}−{b}\right)}{\left({c}−{a}\right)\left({c}−{b}\right)}+\frac{\left({x}−{b}\right)\left({x}−{c}\right)}{\left({a}−{b}\right)\left({a}−{c}\right)}+\frac{\left({x}−{c}\right)\left({x}−{a}\right)}{\left({b}−{c}\right)\left({b}−{a}\right)}=\mathrm{1} \\ $$

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