Question

1/7 (3x-5)^2=28 solve the following equation by factorisation​

Answer 1

꧁ ANSWER ꧂

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$⇒ \frac{1}{7} (3x - 5) {}^{2} = 28 \\ \\ ⇒(3x - 5) {}^{2} = 28 \times 7 \\ \\ ⇒9 {x}^{2} - 30x + 25 = 196 \\ \\ ⇒9 {x}^{2} - 30x + 25 - 196 = 0 \\ \\ ⇒9 {x}^{2} - 30x - 171 = 0 \\ \\ ⇒9 {x}^{2} + 27x - 57x - 171 = 0 \\ \\ ⇒9x(x + 3) - 57(x + 3) = 0 \\ \\ ⇒(9x - 57)(x + 3) = 0 \\ \\ ⇒9x - 57 = 0 \\ \\ ⇒9x = 57 \\ \\ ⇒x = \frac{19}{3} or - 3$

Hence:-

• The value of X=19/3

Answer 2

(3x-5)^2=28*7

9x^2-30x+25-196=0

3(3x^2-10x-57)=0

3x^2-19x+9x-57=0

x(3x-19)+3(3x-19)=0

(x+3)(3x-19)=0

x= -3,19/3