Question

The equation 3(3x^2+15)/6 +2x^2+9=(2x^2+96/7 ) +6 has got the solution as

Answer 1

Answer:

x = -1 & x= 1

Step-by-step explanation:

Root can be positive or negative

Answer 2

The solution is x = ±1

Given

$\frac{3(3x^2 + 15)}{6} + 2x^2 + 9 = 2x^2 + 96/7 + 6$

To Find

The solution for x

Solution

Here we have the equation,

$\frac{3(3x^2 + 15)}{6} + 2x^2 + 9 = 2x^2 + 96/7 + 6$

Expanding the numerator on the LHS and adding the constants up in the RHS we get,

$= \frac{3x^2 + 15}{2} + 2x^2 + 9 = 2x^2 + \frac{96 + 7X6 }{7}$

$or, \frac{3x^2 + 15 + 2X2x^2 + 2X9}{2} = \frac{2X2x^2 + 96 + 42 }{7}$

Simplifying the above result will give us,

$or, \frac{3x^2 + 15 + 4x^2 + 18}{2} = \frac{4x^2 + 136 }{7}$

$or, \frac{7x^2 + 33}{2} = \frac{4x^2 + 136 }{7}$

Now, cross multiplying the denominators in both LHS and RHS, i.e bringing the denominator in RHS i.e 7 to the LHS and that of LHS i.e 2 to the RHS will give us

$or, 7(7x^2 + 33) = 2(4x^2 + 136)$

expanding the above result will give us

$or, 7X7x^2 + 7X33 = 2X4x^2 + 2X136$

$or, 49x^2 + 231 = 8x^2 + 272$

bringing all the x's to the LHS and the constants to RHS we get,

$or, 49x^2 - 8x^2= + 272 - 231$

$or, 41x^2 = 41$

or, x² = 1

or, x = ±1

Therefore, the solution is x = ±1

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