# algebra1

• 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 (x-y) = 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)=(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 -...

• 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 a2-a6+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+m-1 y' = 2mx + 2m + 1 = 0 => 2mx = -2m - 1 => x = (-2m - 1)...

• algebra1
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) =>...

• 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)(x-1) = 24 (3x+7)(x-1) = 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 + (n-1)*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 (x-1), (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=20-x and 3x-2y-6=0. This is equivalent to solving for x and y using the given equations. We have y=20-x. substitute this for y in 3x-2y-6=0 => 3x - 2( 20 -...

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