algebra1

The multiplicative inverse is another name for the reciprocal. Both numbers are such that, if multiplied by the given number, they result in 1. The number 1 is special with regard to the operation...

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 -...

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...

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!...

In this question, I am not clear whether the logs have the base of 3 or 3 is the part of the arguments. I will try to show both cases. Let's consider the given system of equations: 4^(x/y)*4^(y/x)...

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...

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 a2-a6+a4=-7 => a + d - a - 5d + a + 3d = -7 => a - d...

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 =...

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 /...

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 =>...

x^3 - y^3 = 7..........(1) x^2 + xy + y^2 = 7 ..............(2) We need to solve the system. First we will rewrite equation (1) as a difference between cubes. ==> x^3 - y^2 = (x-y)(x^2 +...

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 +...

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...

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 -...

The multiplicative inverse of an element of a set is another element of this set such that the product of the two elements is a multiplicative identity. In the given set of complex numbers, the...

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...

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 -...

We have to find the zeros of the function f(x) = x^2 - 8x - 9 For this we equate f(x) to 0 and solve the quadratic equation that is obtained. f(x) = 0 => x^2 - 8x - 9 = 0 => x^2 - 9x + x - 9...

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...

It is given that log(72) 48 = a and log(6) 24 = b a = log(72) 48 = log(6) 48/ log(6) 72 => log(6) (6*8)/log(6) 6*12 => [1 + log(6) 8]/[1 + log(6) 12] => [1 + 3*log(6) 2]/[2 + log(6) 2] b...

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...

We have to find the complex number z given that (3z - 2z')/6 = -5 Let z = a + ib, z' = a - ib (3z - 2z')/6 = -5 =>(3(a + ib) - 2(a - ib)) = -30 => 3a + 3ib - 2a + 2ib = -30 => a + 5ib =...

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...

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)...

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...

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+m-1 y' = 2mx + 2m + 1 = 0 => 2mx = -2m - 1 => x = (-2m - 1)...

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) =>...

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...

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...

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...

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...

We have to solve: 5(8e^2x - 3)^3 = 625 for x. 5(8e^2x - 3)^3 = 625 divide both sides by 5 => (8e^2x - 3)^3 = 125 take the cube root of both the sides => 8e^2x - 3 = 5 add 3 to both the sides...

We see that for each value of x, f(x)=(2x-5)/(7x+4) has only one value and each value of f(x) can be obtained by only one value of x. The function has an inverse. Let y = f(x)=(2x-5)/(7x+4) express...

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 +...

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...

Consecutive terms of a GP have a common ratio. if 2, x, y, 16 form a GP. => 16/y = x/2 => x = 32/y y/x = x/2 Substitute x = 32/y => y/(32/y) = (32/y)/2 => 2y = (32/y)^2 => 2y^3 =...

We have to determine a and b given that for x*y = xy + 2ax + by, the * operator is commutative. x*y = y*x => xy + 2ax + by = yx + 2ay + bx => 2ax + by = 2ay + bx equating the coefficients of...

The domain of a function f(x) is all the values x for which f(x) gives real values. f(x)=(x-2)/(x^2-4) => (x - 2)/(x - 2)(x + 2) => 1/(x + 2) This is not defined when x = -2 The domain is R -...

It is given that f(x)=14x+13. f(f(4)) given that f(4) = 14*4 + 13 => f( 14*4 + 13) => f(69) => 14*69 + 13 => 979 The value of f(f(4)) = 979

The sum of 3+2i+1-5i is found by adding the real terms together and the complex terms, or the terms that contain i, together. 3+2i+1-5i => 3 + 1 + 2i - 5i => 4 - 3i The required result of...

Two graphs have a point of intersection if the equations of the graph have a solution. The equations of the graphs given are: f(x)=2x+1 and g(x)=x^2+x+1 Equating the two we get 2x + 1 = x^2 + x +...

We have 5x + 4y = 9 and 3x + 2y = 5. We need to find 4x + 3y. Here we don't need to find x and y. The required result can be obtained by adding 5x + 4y = 9 and 3x + 2y = 5 => 5x + 4y + 3x + 2y =...

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

We have to solve x+6 = 6* sqrt (x-2) x+6 = 6* sqrt (x-2) square both the sides (x + 6)^2 = 36( x - 2) => x^2 + 12x + 36 = 36x - 72 => x^2 - 24x + 108 = 0 => x^2 - 18x - 6x + 108 = 0 =>...

As z is a complex number let it be x + yi. z' = x - yi. If z/2 + 3=z'/3 - 2 => (x + yi)/2 + 3 = (x - yi)/3 - 2 => (x + yi)/2 - (x -yi)/3 = -5 => x/2 - x/3 - i( y/2 - y/3) = -5 => x/2 -...

To simplify (8-6i)(-4-4i) open the brackets and multiply the terms (8-6i)(-4-4i) => 8*-4 - 8*4i - 6i*(-4) - 6i*(-4i) => -32 - 32i + 24i + 24i^2 => -32 - 8i - 24 => -56 - 8i The required...

We have to solve 4*|8x - 8| < 32 4*|8x - 8| < 32 => |8x - 8| < 8 => |x - 1| < 1 -1 < (x - 1) < 1 -1 < (x - 1) => 0 < x (x - 1) < 1 => x < 2 The values of x...

We have to solve the system x + y = 3 and x^2/y + y^2/x = 9/2 x + y = 3 => x = 3 - y Substitute this in x^2/y + y^2/x = 9/2 => (3 - y)^2 / y + y^2 / (3 - y) = 9/2 => (3 - y)(3 - y)^2 + y*...

The equation to be solved is 2x^4 - 11x^3 + 23x^2 - 19x + 5 = 0 We can see that for x = 1, 2 - 11 + 23 - 19 + 5 = 0 This gives us one solution as 1. 2x^4 - 11x^3 + 23x^2 - 19x + 5 = 0 => 2x^4 -...

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...