A Treatise on Algebra: By Charles SmithMacmillan and Company, 1888 - 571 sider |
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Side ix
... solving an equation the same as the problem of Quadratic equations . Discussion of roots of a quadratic equation Zero and infinite roots 109 • 109 . 112 • 113 Equations not integral 115 Irrational equations 118 • A quadratic equation ...
... solving an equation the same as the problem of Quadratic equations . Discussion of roots of a quadratic equation Zero and infinite roots 109 • 109 . 112 • 113 Equations not integral 115 Irrational equations 118 • A quadratic equation ...
Side 104
... solve an equation is to find the value or values of the unknown quantity for which the equation is true ; and these values of the unknown quantity are said to satisfy the equation , and are called the roots of the equation . Two ...
... solve an equation is to find the value or values of the unknown quantity for which the equation is true ; and these values of the unknown quantity are said to satisfy the equation , and are called the roots of the equation . Two ...
Side 106
... Solve the equation 13x - 7 = 5x + 9 . Transpose the terms 5x and -7 ; then 13x - 5x = 7 + 9 . That is 8x = 16 . Divide both sides by 8 , the coefficient of x ; then x = 2 . 3x 2x Ex . 2. Solve the equation 2 - = +5 . 4 5 We may get rid ...
... Solve the equation 13x - 7 = 5x + 9 . Transpose the terms 5x and -7 ; then 13x - 5x = 7 + 9 . That is 8x = 16 . Divide both sides by 8 , the coefficient of x ; then x = 2 . 3x 2x Ex . 2. Solve the equation 2 - = +5 . 4 5 We may get rid ...
Side 107
... Solve the equations 1. 1 ( x − 2 ) - 1 ( x − 3 ) + ( x − 4 ) = 4 . - - 4 − 1 Ans . x = 12 . ( x − 8 ) + ( x − 5 ) = 0 . Ans . x = 0 . Ans . x = a + b . 1 3 SIMPLE EQUATIONS . 107 Special forms of simple equations.
... Solve the equations 1. 1 ( x − 2 ) - 1 ( x − 3 ) + ( x − 4 ) = 4 . - - 4 − 1 Ans . x = 12 . ( x − 8 ) + ( x − 5 ) = 0 . Ans . x = 0 . Ans . x = a + b . 1 3 SIMPLE EQUATIONS . 107 Special forms of simple equations.
Side 109
... Solve the equation Transposing , we have that is x _ 5x = 6 . x – 5 −6 = 0 , ( x −6 ) ( x + 1 ) = 0 ; .. x - 6 = 0 , or x + 1 = 0 . Hence x = 6 , or x == - 1 . Ex . 2. Solve the equation Transposing , we have that is x3- x2 - 6x . x3 ...
... Solve the equation Transposing , we have that is x _ 5x = 6 . x – 5 −6 = 0 , ( x −6 ) ( x + 1 ) = 0 ; .. x - 6 = 0 , or x + 1 = 0 . Hence x = 6 , or x == - 1 . Ex . 2. Solve the equation Transposing , we have that is x3- x2 - 6x . x3 ...
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a+b+c a₁ arithmetic mean arithmetical progression ax² b₁ binomial Binomial Theorem C₁ Cambridge coefficient contain continued fraction Crown 8vo denominator determinant digits divergent divide divisible divisor Edition equal equation ax example expansion Extra fcap Fcap Find the number Find the sum follows geometrical progression given expression Globe 8vo greater Hence the required infinity letters logarithm MACMILLAN'S Multiply negative nth convergent nth root nth term number of terms obtained P₁ periodic continued fraction positive integer powers prime number Professor prove quadratic equation quadratic surd quotient ratio remainder School Shew Solve the equation square root subtract suppose Theorem Trinity College u₁ unknown quantities white ball whole number zero
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