
algebra1
We have to solve the equations: y^2 = x^2  9 ...(1) y = x  1 ...(2) From (2) we get y = x  1 => y^2 = (x  1)^2 substitute in (1) (x  1)^2 = x^2  9 => x^2 + 1  2x = x^2  9 => 2x =...

algebra1
We need to solve (log(2) x)^2 + log(2) (4x) = 4 Use the property that log a*b = log a + log b (log(2) x)^2 + log(2) (4x) = 4 => (log(2) x)^2 + log(2) 4 + log(2) x = 4 => (log(2) x)^2 + 2 +...

algebra1
We have the equations: 4^(x/y)*4^(y/x) = 32 ...(1) log 3 (xy) = 1  log 3 (x+y) ...(2) From (1) we get 4^(x/y)*4^(y/x) = 32 => 4^(x/y + y/x) = 2^5 => 2^2(x/y + y/x) = 2^5 equate the...

algebra1
We have the system of equations x^2 + y^2 = 16 and xy = 13 to solve for x. xy = 3 => x = 3/y...(1) Substitute this in x^2 + y^2 = 16 => (3/y)^2 + y^2 = 16 => 9 + y^4 = 16y^2 => y^4 ...

algebra1
We have to solve the simultaneous equations x^2 + y^2=10 ...(1) x^4 + y^4=82 ...(2) let a = x^2 and b = y^2 This gives a + b = 10 and a^2 + b^2 = 82 a + b = 10 => a^2 + b^2 + 2ab = 100...

algebra1
A polynomial has complex roots in pairs of conjugates. As the polynomial has roots 2 and 2i, it also has 2i as a root. The polynomial is: (x  2)(x  2i)(x + 2i) => (x  2)(x^2  4i^2) => (x...

algebra1
The domain of a function f(x) is all the values x for which f(x) gives real values. f(x)=(x2)/(x^24) => (x  2)/(x  2)(x + 2) => 1/(x + 2) This is not defined when x = 2 The domain is R ...

algebra1
We have to find the real solutions of log(x+2) x + log(x) (x+2) = 5/2. We use the relation log (a) b = 1/log (b) a and log a + log b = log a*b log(x+2) x + log(x) (x+2) = 5/2 => 1/ log (x) (x+2)...

algebra1
The multiplicative inverse of a term x is given by y such that x*y = 1. Now we have to find the multiplicative inverse of 6  3i Let it be x + yi (6  3i)(x + yi) = 1 => x + yi = 1/(6  3i)...

algebra1
We have to solve the following set of simultaneous equations: 2x  3y = 5 ...(1) x  2y = 4 ...(2) From (2) x  2y = 4 => x= 4 + 2y Substitute in (1) 2( 4 + 2y)  3y = 5 => 8 + 4y  3y = 5...

algebra1
The roots of x^3  9x^2 + 23x  15 = 0 are in AP. x^3  9x^2 + 23x  15 = 0 => x^3  3x^2  6x^2 + 18x + 5x  15 = 0 => x^2(x 3 )  6x ( x  3) + 5(x  3) =0 => (x^2  6x + 5)(x  3) = 0...

algebra1
The nth term of an arithmetic sequence can be written as a + (n  1)*d, where a is the first term and d is the common difference. We have a2a6+a4=7 => a + d  a  5d + a + 3d = 7 => a  d...

algebra1
We have to solve for x given that x^log2 x + 8*x^log2 x = 6 x^log(2) x + 8*x^( log(2) x) = 6 => x^log(2) x + 8/x^(log(2) x) = 6 Let x^log(2) x = y => y + 8/y = 6 => y^2  6x + 8 = 0...

algebra1
We have to solve the equation sqrt (x + sqrt (1  x)) + sqrt x = 1 sqrt (x + sqrt (1  x)) + sqrt x = 1 sqrt (x + sqrt (1  x)) = 1  sqrt x square both the sides => x + sqrt(1  x) = 1 + x  2...

algebra1
We have x^2 + y^2 = 100 and x  y = 2. We have to solve these equations for x and y. x  y = 2 => (x  y)^2 =2^2 => x^2 + y^2  2xy = 4 substitute x^2 + y^2 = 100 => 100  2xy = 4 =>...

algebra1
The multiplicative inverse of a number A is another number B which when multiplied by A gives 1. Here we have the number 3 + 2i Let A = 3 + 2i, we need to find B such that A*B = 1 => B = 1/(3 +...

algebra1
To solve 2x  y = 5 ...(1) 3x  2y = 9 ...(2) using substitution take (1) 2x  y = 5 => y = 2x  5 substitute in (2) 3x  2(2x  5) = 9 => 3x  4x + 10 = 9 => x = 1 => x = 1 y = 2x ...

algebra1
The roots of a quadratic equation are given as 6 and 7. This means that x  6 = 0 and x  7 = 0 We can write ( x  6)( x  7) = 0 => x^2  6x  7x + 42 = 0 => x^2  13x + 42 = 0 Therefore the...

algebra1
The roots of the quadratic equation are x1 and x2. x1  x2 = 1 and x^2 = 2x  m As x^2 = 2x  m x1^2 = 2* x1  m x2^2 = 2*x2  m Subtracting the two we get x1^2  x2^2 = 2( x1  x2) => (x1 ...

algebra1
We have to determine the imaginary part of the complex number z for z^2 = 3 + 4i Let z = x + yi z^2 = x^2 + y^2*i^2 + 2xyi = 3+ 4i => x^2  y^2 + 2xyi = 3 + 4i equate the real and imaginary...

algebra1
We have the system of equations: x^3  y^3 = 7 x^2 + xy + y^2 = 7 x^3  y^3 = 7 => (x  y)(x^2 + xy + y^2) = 7 => (x  y) *7 = 7 => x  y = 1 = x = 1 + y x^2 + xy + y^2 = 7 => (1 +...

algebra1
k! = 1*2*3*...*k (k  1)/k! = k/k!  1/k! => 1/(k  1)!  1/k! The sum of (k  1)/k! for k = 1 to n is: 1/0!  1/1! + 1/1!  1/2! + 1/2!  1/3! + ... + 1/(n  1)!  1/n! => 1/0!  1/n!...

algebra1
We have to solve for x: log2 x + log4 x + log8 x =11/6 log (a) b = 1/ log (b) a log2 x + log4 x + log8 x =11/6 => 1 / log(x) 2 + 1 / log(x) 4 + 1/log(x) 2 = 11/6 => 1 / log(x) 2 + 1 /...

algebra1
We have to solve: e^2x  6e^x + 5 = 0 e^2x  6e^x + 5 = 0 let e^x = t => t^2  6t + 5 = 0 => t^2  5t  t + 5 = 0 => t(t  5)  1( t  5) = 0 => (t  1)(t  5) = 0 => e^x = 1 and e^x...

algebra1
We have to solve 2x  y = 9 ...(1) 5x  2y = 18 ...(2) 2*(1)  (2) => 4x  2y  5x + 2y = 18  18 => 9x = 36 => x = 4 substitute in (1) 2x  y = 9 => y = 2*4 + 9 => y = 1...

algebra1
Let the first term of the arithmetic sequence be a and the common difference be d. (a + a + 12d)*(13/2) = 130 => 2a + 12d = 20 => a + 6d = 10 => a = 10  6d a4, a10 and a7 are consecutive...

algebra1
We are given that f(x)= 6^x and g(x) = log(6) x We have to find fog(36) fog(36) = f(g(36)) => f(log(6) 36) => 6^(log (6) 36) we know that a^(log(a) x) = x => 36 The required solution for...

algebra1
The polar form of a complex number z = x + yi is r*(cos A + i* sin A) where tan A= y/x and r = sqrt (x^2 + y^2) Here we have z = 6  8i r = sqrt( 6^2 + 8^2) = sqrt (36 + 64) = sqrt 100 = 10 tan A =...

algebra1
We have the equation: 2x^3  x^2 + ax + b = 0 with one of its solutions given as 1 + i. Substituting x with 1 + i, 2*(1 + i)^3  (1 + i)^2 + a(1 + i) + b = 0 => 2*( 1 + 3i^2 + 3i + i^3)  1 ...

algebra1
The terms x + 4, 3x and 13x  2 are consecutive terms of an arithmetic progression. So we know that they have a common difference 13x  2  3x = 3x  x  4 => 10x  2 = 2x  4 => 8x = 2...

algebra1
We have to solve 2z + 6i = z/2i + 5i  7 2z + 6i = z/2i + 5i  7 => 2z  z/ 2i = i  7 => (2z*2i  z) / 2i = (i  7) => z( 4i  1)/2i = (i  7) => z = (i  7) * 2i / ( 4i  1) =>...

algebra1
We can write 7.2 as 6*6/5 log 7.2 = log 6*6 / 5 = log 6^2 / 5 Now log (a^b) = b* log a and log (a/b) = log a  log b log (6^2 / 5) => 2 log 6  log 5 as log 6 = m and log 5 = n => 2m  n...

algebra1
We are given that y = 3x / (x^2  9). We have to determine m and n if y = m/(x  3) + n/(x +3) Equate the two expressions for y. 3x / (x^2  9) = m/(x  3) + n/(x +3) => 3x / (x^2  9) = [m(x +...

algebra1
The extreme of the parabola is determined by finding the first derivative and equating it to 0. We then solve for x. y = mx^2+2mx+x+m1 y' = 2mx + 2m + 1 = 0 => 2mx = 2m  1 => x = (2m  1)...

algebra1
The complex number (2+2i)^11/(22i)^9 has to be simplified first. (2+2i)^11/(22i)^9 => (4 + 4i^2 + 8i)^5*(2 + 2i) / (4 +4i^2  8i)^4 * (2  2i) => (8i)^5*(2 + 2i) / ( 8i)^4 * (2  2i) =>...

algebra1
The inverse of the function f(x) = 5*tan(3x+4) is required. Let y = f(x) = 5*tan(3x+4) Now express x in terms of y => tan (3x + 4) = y/5 => 3x + 4 = arc tan (y/5) => 3x = arc tan (y/5) ...

algebra1
The equation to be solved is : (3x+7)(x1) = 24 (3x+7)(x1) = 24 => 3x^2 + 4x  7 = 24 => 3x^2 + 4x  31 = 0 x1 = 4/6 + sqrt (16 + 372) /6 => 2/3 + (sqrt 97)/6 x2 = 2/3  (sqrt 97)/6...

algebra1
We have to solve the following set of simultaneous equations: x + 3y = 2 ...(1) 2x + 8y = 6 ...(2) 2*(1)  (2) => 2x + 6y  2x  8y = 4  6 => 2y = 2 => y = 2/2 => y = 1 substitute...

algebra1
We have to verify if (a+b)^5  a^5  b^5 = 5ab(a+b)(a^2+ab+b^2) We can expand (a + b)^5 = a^5 + 5a^4b + 10 a^3b^2 + 10 a^2b^3 + 5ab^4 + b^5 The left hand side (a+b)^5  a^5  b^5 => a^5 + 5a^4b...

algebra1
For an AP the nth terms can be written as a + (n1)*d, where a is the first term and d is the common difference between consecutive terms. The sum of the first n terms is (t1 + tn)*(n/2) In the...

algebra1
We have to solve: x + y  14 = 0 ...(1) 4x  y  11 = 0 ...(2) From (1) x + y  14 = 0 => y = x  14 Substitute in (2) 4x + x  14  11 = 0 => 5x  25 = 0 => x = 25/5 => x = 5 y = 14 ...

algebra1
The polynomial ax^3 + bx^2 + cx + d is the product of the terms (x1), (x+1) and (x+2) to which the remainder 3 is added. (x  1)(x +1)(x + 2) +3 => (x^2  1)(x + 2) +3 => x^3  x + 2x^2  2...

algebra1
We have to factor a^4  b^16 we use x^2  y^2 = (x +y)(x  y) a^4  b^16 => (a^2  b^8)(a^2 + b^8) => (a  b^4)(a + b^4)(a^2 + b^8) Therefore a^4  b^16 = (a  b^4)(a + b^4)(a^2 + b^8)

algebra1
We have to solve (x+57)/6=x^2 (x+57)/6=x^2 => x + 57 = 6x^2 => 6x^2  x  57 = 0 The roots of a quadratic equation ax^2 + bx + c = 0 are given by [b + sqrt (b^2  4ac)]/2a and [b + sqrt...

algebra1
We have to solve for x given 64x^3+1331=0 64x^3 + 1331 = 0 => (4x)^3 + 11^3 = 0 => (4x + 11)( (4x)^2  4x*11 + 11^2) = 0 4x + 11 = 0 => x1 = 11/4 => x = 2.75 ( (4x)^2  4x*11 + 11^2)...

algebra1
The equation you have given is z^4  3z^2 + 2 = 0. Here you can replace z^2 = x to get a quadratic equation in x which can be solved. z^4  3z^2 + 2 = 0 => x^2  3x + 2 = 0 => x^2  2x  x +...

algebra1
We have to find the common points of y=20x and 3x2y6=0. This is equivalent to solving for x and y using the given equations. We have y=20x. substitute this for y in 3x2y6=0 => 3x  2( 20 ...

algebra1
We have the equation 11^(5x6)=1/11^(x10) to solve for x. 11^(5x6)=1/11^(x10) => 11^(5x  6) = 11^[(  x  10)] => 11^(5x  6) = 11^(x + 10) as the base is the same we can equate the...

algebra1
We have to solve (x+5)^24(4x+5)=0 (x+5)^2  4(4x+5) = 0 open the brackets x^2 + 10x + 25  16x  20 = 0 => x^2  6x + 5 = 0 => x^2  5x  x + 5 = 0 => x( x  5)  1(x  5) = 0 => (x ...

algebra1
We have to solve the equations 1/x=(y1)/y and 3/x=(2y1)/y 1/x=(y1)/y => x = y/(y  1) Substitute x = y/ (y  1) in 3/x=(2y1)/y => 3/x=(2y1)/y => 3 /[y/ (y  1)] = (2y1)/y => 3(y ...