Question

(2x+3)(2x+5)(x-1)(x-2)=30​

Answer 1

Answer:

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Step-by-step explanation:

$to \: find = \\ values \: of \: x \\ \\ so \: here \: \\ (2x + 3)(2x + 5)(x - 1)(x - 2) = 30 \\ \\ (2x + 3)(x - 1)(2x + 5)(x - 2) = 30 \\ \\ (2x {}^{2} - 2x + 3x - 3)(2x {}^{2} - 4x + 5x - 10) = 30 \\ \\ (2x {}^{2} + x - 3)(2x {}^{2} + x - 10) = 30 \\ \\ so \: let \\ 2x {}^{2} + x = k \\ hence \: accordingly \\ \\ (k - 3)(k - 10) = 30 \\ k {}^{2} - 10k - 3k + 30= 30 \\ k {}^{2} - 13k = 0 \\ ie \: \: k(k - 13) = 0 \\ so \: \: k = 0 \: \: or \: \: k - 13 = 0 \\ \\ k = 0 \: \: or \: \: k = 13$

$so \: resubstituting \: value \: of \: k \: firstly \: as \: 0\: \\ we \: get \\ 2x {}^{2} + x = 0 \\ x(2x + 1) = 0 \\ \\ x = 0 \: \: or \: \: 2x + 1 = 0 \\ \\ ie \: x = 0 \: \: or \: \: x = \frac{ - 1}{2}$

$now \: resubstituting \: value \: of \: k \: as \: 13 \\ we \: get \\ 2x {}^{2} + x = 13 \\ \\ 2x {}^{2} + x - 13 = 0 \\ comparing \: this \: quadratic \: equation \: with \\ ax {}^{2} + bx + c = 0 \\ we \: get \\ a = 2 \: \: b = 1 \: \: c = - 13 \\ \\ now \: using \\ Δ = b {}^{2} - 4ac \\ = (1) {}^{2} - 4 \times 2 \times ( - 13) \\ = 1 - 8( - 13) \\ = 1 - ( - 104) \\ = 1 + 104 \\ = 105 \\ \\ so \: by \: applying \: formula \: method \\ we \: get \\ \\ x = \frac{ - b \:± \: \sqrt{b {}^{2} - 4ac } }{2a} \\ \\ = \frac{ - 1 \:± \: \sqrt{105} }{2 \times 2} \\ \\ \: x = \frac{ - 1 \: + \: \sqrt{105} }{4} \: \: \: \: or \: \: \: x = \frac{ - 1 \: - \: \sqrt{105} }{4}$

Answer 2

Answer:

(2x + 3) (2x + 5)(x – 1)(x – 2) = 30. {(2x + 3)(x – 1)}{(2x + 5)(x – 2)} = 30. (2x2 – 2x + 3x – 3)(2x2 – 4x + 5x – 10) = 30.

Explanation: