Question

Solve 1/7(3x-5)^2=28 by factorisation

Answer 1

Answer:

$\frac{1}{7} (3x - 5) {}^{2} = 28 \\ \frac{1}{7} (9x {}^{2} + 25 - 30x) = 28 \\ 9x {}^{2} - 30x + 25 = 196 \\ 9x {}^{2} - 30x + 25 - 196 = 0 \\9 x {}^{2} - 30x - 171 = 0$

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Answer 2

$\frac{1}{7} (3x-5)^{2} = 28$

/* Multiplying bothsides by 7 , we get */

$\implies (3x-5)^{2} = 7 \times 28$

$\implies (3x-5)^{2} - 196 = 0$

$\implies (3x-5)^{2} - 14^{2} = 0$

$\implies (3x-5+14)(3x-5-14) = 0$

$\implies (3x+9)(3x-19) = 0$

$\implies 3x + 9 = 0\: Or \:3x-19 = 0$

$\implies 3x = -9 \: Or \: 3x = 19$

$\implies x = -3\: Or \: x = \frac{19}{3}$

Therefore.,

$\green { x = -3\: Or \: x = \frac{19}{3}}$

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