A Trending Radical Equation With Square Roots.

How solve this trending radical equation.

[GillesF]

I did it this way in order to avoid (to try!!!) to make as few errors as possible in my calculations:

√(x + 4) + √(2x - 1) = 3√(x - 1)

set:

• a = (x + 4)

• b = (2x - 1)

• c = (x - 1)

then √(x + 4) + √(2x - 1) = 3√(x - 1) becomes:

√a + √b = 3√c

(√a + √b)² = (3√c)²

a + 2√(ab) + b = 9c

2√(ab) = 9c - a - b

(2√(ab)² = (9c - a - b)²

4ab = (9c - a - b)·(9c - a - b)

4ab = 81c² - 9ac - 9bc - 9ac + a² + ab - 9bc + ab + b²

81c² - 9ac - 9bc - 9ac + a² + ab - 9bc + ab + b² - 4ab = 0

a² + b² + 81c² - 9ac - 9ac - 9bc - 9bc + ab + ab - 4ab = 0

a² + b² + 81c² - 18ac - 18bc - 2ab = 0

(x + 4)² + (2x - 1)² + 81(x - 1)² - 18(x + 4)(x - 1) - 18(2x - 1)(x - 1) - 2(x + 4)(2x - 1) = 0

||
|| (x + 4)² = x² + 8x + 16
||
|| (2x - 1)² = 4x² - 4x + 1
||
|| 81(x - 1)² = 81(x² - 2x + 1) = 81x² - 162x + 81
||
|| 18(x + 4)(x - 1) = 18(x² + 3x - 4) = 18x² + 54x - 72
||
|| 18(2x - 1)(x - 1) = 18(2x² - 3x + 1) = 36x² - 54x + 18
||
|| 2(x + 4)(2x - 1) = 2(2x² + 7x - 4) = 4x² + 14x - 8
||

(x² + 8x + 16) + (4x² - 4x + 1) + (81x² - 162x + 81) - (18x² + 54x - 72) - (36x² - 54x + 18) - (4x² + 14x - 8) = 0

x² + 8x + 16 + 4x² - 4x + 1 + 81x² - 162x + 81 - 18x² - 54x + 72 - 36x² + 54x - 18 - 4x² - 14x + 8 = 0

x² + 4x² + 81x² - 18x² - 36x² - 4x² + 8x - 4x - 162x - 54x + 54x - 14x + 16 + 1 + 81 + 72 - 18 + 8 = 0

28x² - 172x + 160 = 0

(28x² - 172x + 160)/4 = 0/4

7x² - 43x + 40 = 0

Δ = (-43)² - 4·7·40 = 1849 - 1120 = 729

√Δ = ±√729= ±27

• root #1: x = (-(-43) + 27)/(2·7) = (43 + 27)/14 = 70/14 = 5

• root #2: x = (-(-43) - 27)/(2·7) = (43 - 27)/14 = 16/14 = 8/7

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/// check:

// x = 5

√(5 + 4) + √(2·5 - 1) - 3√(5 - 1) = 0 => x = 5 is a solution

// x = 8/7

√(8/7+ 4) + √(2·(8/7) - 1) - 3√(8/7 - 1) ≠ 0 => x = 8/7 isn't a solution

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/// final result:

■ x = 5

🙂