Find the possible dimensions for the cuboids whose volume is: V=6kt^2 - 21kt - 12k  It is based on the chapter polynomials 9th standard CBSE(NCERT) syllabus.

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sciencesolve | Teacher | (Level 3) Educator Emeritus

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You should use the equation that gives the volume of cuboid such that:

`V = L*w*h`

`L, w`  represent the length and width of the base of cuboid and `h`  represents the height of cuboid

The problem provides the quadratic equation that represents the volume of cuboid such that:

`V(t) = 6kt^2 - 21kt - 12k`

Notice that `6k` , `-21k`  and `-12 k`  represents the coefficients of the quadratic equation.

You nedd to convert the standard form of quadratic into the factored form such that:

`V(t) = 6k(t - t_1)(t - t_2)`

`6k, t - t_1, t - t_2`  represent the three dimensions of cuboid

`t_(1,2)`  represent the roots of quadratic

You need to use quadratic formula to find `t_1`  and `t_2`  such that:

`t_(1,2) = (21k +- sqrt(441k^2 + 288k^2))/(12k)`

`t_(1,2) = (21k +- sqrt(729k^2))/(12k)`

`t_(1,2) = (21k +- 27k)/(12k)`

`t_1 = 4k ; t_2 = -k/2`

`V(t) = 6k(t - 4k)(t + k/2)`

Hence, evaluating the dimensions of the given cuboid, under the given conditions, yields `t + k/2, t - 4k`  and `6k` .

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vaaruni | High School Teacher | (Level 1) Salutatorian

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Given :    V= 6kt^2 - 21kt - 12k

       =>  V = 3k(2t^2 - 7t - 4)

       => V = 3k( 2t^2- 8t + t - 4)

       => V = 3k{2t(t-4) + 1(t-4)}

       => V = 3k(t-4)(2t+1)

    As we know that volume of a cuboid (V) = length *width *height

 and the  Volume of the cuboid is given as V = 6kt^2 -21kt -12k

which is simplied as V = 3k *(t-4) *(2t+1)  Therefore,

we conclude the dimensions of the cuboid to be 3k, (t-4) and (2t+1).

Possible dimensions of the cuboid : 3k, (t-4) and (2t+1)<-Answer

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