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प्रश्न
Solve: (2x – 1)(x + 3)(x – 2)(2x + 3) + 20 = 0
उत्तर
(x + 3)(x – 2)(2x – 1)(2x + 3) + 20 = 0
(x2 + x – 6)(4x2 + 6x – 2x – 3) + 20 = 0
(x2 + x – 6)(4x2 + 4x – 3) + 20 = 0
`(x^2 + x - 6)(x^2 + 4x - 3/4) + 20` = 0
`(÷ 4)(x^2 + x - 6)(x^2 + x - 3/4) + 20/4` = 0
Put y = x2 + x
`(y - 6)(y - 3/4) + 20/4` = 0
`y^2 - 3/4 y - 6y + 9/2 + 20/4` = 0
`y^2 - y (27/4) + 38/4` = 0
4y2 – 27y – 38 = 0
4y2 – 8y – 19y + 38 = 0
(4y – 19)(y – 2) = 0
y = 2 or y = `19/4`
`x^2 + x = 2` or `x^2 + x = 19/4`
x2 + x – 2 = 0 or 4x2 + 4x = 19
(x + 2)(x – 1) = 0 or 4x2 + 4x – 19 = 0
x = – 2 or x = 1
x = `( - 4 +- sqrt(16 - 4(4)(- 19)))/(2(4)`
= `(- 4 +- sqrt(16 + 304))/(2(4))`
= `(- 4 +- sqrt(64 xx 5))/(2(4))`
= `(- 4 +- 8 sqrt(5))/(2(4))`
= `(4(- 1 +- 2sqrt(5)))/(2(4))`
= `(-1 +- 2sqrt(5))/2`
Roots are `1, -2, (- 1 +- 2sqrt(5))/2`
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