(X+1) (X+2) (X+3) (X+4) = 120

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(X+1) (X+2) (X+3) (X+4) = 120

Posted by Nitesh Ganjhu 4 years, 9 months ago

CBSE > Class 10 > Mathematics

2 answers

Rajneesh Payal 4 years, 9 months ago

Rewrite the equation as: (x + 1)(x + 4)(x + 3)(x + 2) = 120 Multiply the first 2 and last 2 expressions: ..... (1) Let (1) becomes, (y + 4)(y + 6) = 120 y2 + 10x + 24 = 120 y2 + 10x - 96 = 0 y2 + 16x - 6y - 96 = 0 y(y + 16) - 6(y + 16) = 0 (y + 16)(y - 6) = 0 y = -16 or 6 y2+ 5x = 16 or 6 When x2 + 5x = 6 x2 + 5x - 6 = 0 x2 + 6x - x - 6 = 0 x(x + 6) - 1(x + 6) = 0 (x + 6)(x - 1) = 0 x = -6, 1 When x^2 + 5x = -16 x^2 + 5x + 16 = 0 Using quadratic These are all the possible solutions. But the only real solutions are -6 and 1.

0Thank You

Yogita Ingle 4 years, 9 months ago

Rewrite the equation as:

(x + 1)(x + 4)(x + 3)(x + 2) = 120

Multiply the first 2 and last 2 expressions:

{tex}(x^2 + 5x + 4)(x^2 + 5x + 6) = 120{/tex} ..... (1)

Let {tex}x^2 + 5x = y{/tex}

(1) becomes,

(y + 4)(y + 6) = 120

y² + 10x + 24 = 120

y² + 10x - 96 = 0

y² + 16x - 6y - 96 = 0

y(y + 16) - 6(y + 16) = 0

(y + 16)(y - 6) = 0

y = -16 or 6

y²+ 5x = 16 or 6

When x² + 5x = 6

x² + 5x - 6 = 0

x² + 6x - x - 6 = 0

x(x + 6) - 1(x + 6) = 0

(x + 6)(x - 1) = 0

x = -6, 1

When x^2 + 5x = -16

x^2 + 5x + 16 = 0

Using quadratic formula:

{tex}x = [-5 +/- sqrt (25 - 64)]/2{/tex}

The solutions are, -6, 1, [-5 - sqrt (25 - 64)]/2 and [-5 + sqrt (25 - 64)]/2

These are all the possible solutions. But the only real solutions are -6 and 1.

0Thank You

ANSWER

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