A Treatise on Algebra: Containing the Latest Improvements. Adapted to the Use of Schools and CollegesHarper & Brothers, 1846 - 503 sider |
Inni boken
Resultat 1-5 av 14
Side x
... Cubic Equation 449 Resolution of the Equation of the fourth Degree Irrational Expressions analogous to those obtained in the Resolution of Equations of the third Degree 452 . 455 THE DIOPHANTINE ANALYSIS . Introductory Remarks Examples ...
... Cubic Equation 449 Resolution of the Equation of the fourth Degree Irrational Expressions analogous to those obtained in the Resolution of Equations of the third Degree 452 . 455 THE DIOPHANTINE ANALYSIS . Introductory Remarks Examples ...
Side 130
... equation is called an identical equation . 4. Finally , the equation may ... cubic equations , or equations of the third degree ; biquad- ratic equations ... equations depends upon the degree of 130 ALGEBRA . SIMPLE EQUATIONS.
... equation is called an identical equation . 4. Finally , the equation may ... cubic equations , or equations of the third degree ; biquad- ratic equations ... equations depends upon the degree of 130 ALGEBRA . SIMPLE EQUATIONS.
Side 309
... equation will as- sume the form ( x - a1 ) ( x - a ) ( x - a3 ) ..... ( x - a , ) = 0 ; and , consequently , there ... cubic equation x3 - 6x2 + 11r - 6 = 0 is 1 ; find the quadratic containing the other roots . Ans . x - 5x + 6 = 0 ( 4 ) ...
... equation will as- sume the form ( x - a1 ) ( x - a ) ( x - a3 ) ..... ( x - a , ) = 0 ; and , consequently , there ... cubic equation x3 - 6x2 + 11r - 6 = 0 is 1 ; find the quadratic containing the other roots . Ans . x - 5x + 6 = 0 ( 4 ) ...
Side 328
... equation , and the situation of the roots will be indicated by the signs arising from the sub- stitution of the ... cubic equation x3 + ax2 + bx + c = 0 , gives the fol lowing functions , viz .: With the second term . V = x2 + ax2 + bx + ...
... equation , and the situation of the roots will be indicated by the signs arising from the sub- stitution of the ... cubic equation x3 + ax2 + bx + c = 0 , gives the fol lowing functions , viz .: With the second term . V = x2 + ax2 + bx + ...
Side 329
... cubic equation . Next , to find the situation of the roots we must employ narrower limits than + ∞ and -∞ . ∞ . Commencing at zero , let us extend the limits both ways , and , since the proposed equation has only one permanence of ...
... cubic equation . Next , to find the situation of the roots we must employ narrower limits than + ∞ and -∞ . ∞ . Commencing at zero , let us extend the limits both ways , and , since the proposed equation has only one permanence of ...
Innhold
65 | |
77 | |
83 | |
89 | |
97 | |
100 | |
107 | |
117 | |
119 | |
128 | |
134 | |
139 | |
142 | |
155 | |
172 | |
173 | |
181 | |
191 | |
199 | |
205 | |
218 | |
224 | |
232 | |
239 | |
292 | |
298 | |
302 | |
319 | |
361 | |
390 | |
397 | |
403 | |
409 | |
415 | |
422 | |
428 | |
430 | |
433 | |
439 | |
447 | |
455 | |
467 | |
471 | |
473 | |
479 | |
486 | |
493 | |
501 | |
Andre utgaver - Vis alle
A Treatise on Algebra: Containing the Latest Improvements. Adapted to the ... Charles William Hackley Uten tilgangsbegrensning - 1846 |
A Treatise on Algebra, Containing the Latest Improvements Charles William Hackley Uten tilgangsbegrensning - 1850 |
A Treatise on Algebra: Containing the Latest Improvements. Adapted to the ... Charles William Hackley Uten tilgangsbegrensning - 1847 |
Vanlige uttrykk og setninger
algebraic arithmetical arithmetical progression becomes binomial binomial theorem called change the signs coefficients column common denominator common divisor Completing the square consequently courier cube root decimal difference divide dividend division elimination equa EXAMPLE exponent expression Extract the square figures formula functions geometrical progression give given equation given number greater greatest common divisor greatest common measure Hence imaginary roots indeterminate inequation infinite least common multiple letters logarithm manner method modulus monomial multiplied negative nth power nth root number of terms number of variations obtain perfect square permutations polynomial present value problem proportion proposed equation quadratic quadratic equation quotient radical ratio real roots reduce remainder represent result rule second term solution square root substituting subtract successive suppose theorem third tion Transposing unknown quantity V₁ Whence whole number
Populære avsnitt
Side 76 - Multiply the divisor thus increased, by the second term of the root, and subtract the product from the remainder.
Side 23 - Divide the first term of the dividend by the first term of the divisor, and write the result as the first term of the quotient. Multiply the whole divisor by the first term of the quotient, and subtract the product from the dividend.
Side 237 - B set out from two towns, which were distant 247 miles, and travelled the direct road till they met. A went 9 miles a day ; and the number of days, at the end of which they met, was greater by 3 than the number of miles which B went in a day. How many miles did each go ? 17.
Side 261 - The logarithm of any power of a number is equal to the logarithm of the number multiplied by the exponent of the power.
Side 122 - Proportion is an equality of ratios. Thus, if a, b, c, d are four quantities, such that a, when divided by b, gives the same quotient as c when divided by d, then a, b, c, d are called proportionals, and we say that a is to b as c is to d...
Side 107 - There will be as many figures in the root as there are periods in the given number.
Side 236 - A's journey. How far did each travel ? A 72 miles. B 54 miles. 9. A company at a tavern had £8 15s. to pay for their reckoning ; but before the bill was settled, two of them left the room, and then those who remained had 10s. apiece more to pay than before : how many were there in the company ? Ans. 7.
Side 237 - There are two square buildings, that are paved with stones, a foot square each. The side of one building exceeds that of the other by 12 feet, and both their pavements taken together contain 2120 stones. What are the lengths of them separately ? Ans.
Side 69 - To divide powers of the same base, subtract the exponent of the divisor from the exponent of the dividend.
Side 49 - Multiply all the numerators together for a new numerator, and all the denominators together for a new denominator.