What are the solutions to the equation solvefor a,(a+b+c+d)^2=4(a+b)(c+d) ?

The solutions to the equation solvefor a,(a+b+c+d)^2=4(a+b)(c+d) are a=-b+c+d

Find the zeros of solvefor a,(a+b+c+d)^2=4(a+b)(c+d)

The zeros of solvefor a,(a+b+c+d)^2=4(a+b)(c+d) are a=-b+c+d

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solve for

a, (a + b + c + d)^{2} = 4 (a + b) (c + d)

a = - b + c + d

Solution steps

(a + b + c + d)^{2} = 4 (a + b) (c + d)

Expand

(a + b + c + d)^{2} : 2 ab + 2 b c + 2 b d + b^{2} + a^{2} + 2 a c + c^{2} + 2 a d + 2 c d + d^{2}

Expand

4 (a + b) (c + d) : 4 a c + 4 a d + 4 b c + 4 b d

2 ab + 2 b c + 2 b d + b^{2} + a^{2} + 2 a c + c^{2} + 2 a d + 2 c d + d^{2} = 4 a c + 4 a d + 4 b c + 4 b d

Subtract

4 b c + 4 b d

from both sides

2 ab + 2 b c + 2 b d + b^{2} + a^{2} + 2 a c + c^{2} + 2 a d + 2 c d + d^{2} - (4 b c + 4 b d) = 4 a c + 4 a d + 4 b c + 4 b d - (4 b c + 4 b d)

Simplify

b^{2} + 2 ab - 2 b c - 2 b d + a^{2} + c^{2} + 2 a c + d^{2} + 2 a d + 2 c d = 4 a c + 4 a d

Subtract

4 a c + 4 a d

from both sides

b^{2} + 2 ab - 2 b c - 2 b d + a^{2} + c^{2} + 2 a c + d^{2} + 2 a d + 2 c d - (4 a c + 4 a d) = 4 a c + 4 a d - (4 a c + 4 a d)

Simplify

a^{2} + (2 b - 2 c - 2 d) a + b^{2} + c^{2} - 2 b c + d^{2} - 2 b d + 2 c d = 0

Solve with the quadratic formula

a_{1, 2} = \frac{- ( 2 b - 2 c - 2 d ) \pm ( 2 b - 2 c - 2 d ) ^{2} - 4 \cdot 1 \cdot ( b ^{2} + c ^{2} - 2 b c + d ^{2} - 2 b d + 2 c d )}{2 \cdot 1}

(2 b - 2 c - 2 d)^{2} - 4 \cdot 1 \cdot (b^{2} + c^{2} - 2 b c + d^{2} - 2 b d + 2 c d) = 0

a_{1, 2} = \frac{- ( 2 b - 2 c - 2 d ) \pm 0}{2 \cdot 1}

a = \frac{- ( 2 b - 2 c - 2 d )}{2 \cdot 1}

Simplify

\frac{- ( 2 b - 2 c - 2 d )}{2 \cdot 1} : - b + c + d

a = - b + c + d

The solution to the quadratic equation is:

a = - b + c + d

x^{2} - 4 x - 24 = - 9 x

x^{2} - 3 x - 38 = - 10

\frac{4 x}{3} - \frac{1}{3} x (x - 1) = x^{2} + x

x^{2} + 12 x - 253 = 0

(2 x - 1)^{2} - (x + 1)^{2} = 9

What are the solutions to the equation solvefor a,(a+b+c+d)^2=4(a+b)(c+d) ?
The solutions to the equation solvefor a,(a+b+c+d)^2=4(a+b)(c+d) are a=-b+c+d
Find the zeros of solvefor a,(a+b+c+d)^2=4(a+b)(c+d)
The zeros of solvefor a,(a+b+c+d)^2=4(a+b)(c+d) are a=-b+c+d