Solving a 'Harvard' University entrance exam |Find k?

[00:00:00]Hello everyone. Welcome to solve this nice math olympiad algebra problem. So here we have k times of square root of k is equals to 8. And we solve this problem for all the values of k.

[00:00:14]So now we move towards the solution of this problem.

[00:00:17]Here our target is to remove this square root sign. And in order to remove the square root sign, we need to take squaring on both of the sides. So when we apply squaring, this will be written as k times of square root of k, and its whole square is equals to 8 squared.

[00:00:38]Because we take squaring on both of the sides.

[00:00:41]And now we use here the nice square rule. According to this rule, we apply the square separately on both of these two values, and we get k squared into square root of k and its whole square is equals to this is 8 squared, and we know that 8 squared equals to 64.

[00:01:04]So further in the next step, we have this is k squared times this two and this square root are gone.

[00:01:12]And we get here uh from here we get k only, and this is equals to 64.

[00:01:21]So here we multiply k squared with k, and we get k cubed is equals to 64.

[00:01:30]And now we move the term 64 from right-hand side to the left-hand side, and this will becomes k cubed minus 64.

[00:01:39]And this whole equation is equals to zero.

[00:01:43]And now here we rewrite this equation as k cubed minus this is 64.

[00:01:50]And we write it as 4 cubed, and this whole equation is equals to zero.

[00:01:56]So here we use the nice cubic formula to solve this equation. So, we know about that if we have a cubed minus b cubed equals to a minus b times of a squared plus a b plus b squared.

[00:02:16]So, according to this nice rule we expand the left-hand side of the above equation as this will becomes a minus b. Here, the value of a is k minus the value of b is four into this is a squared. The value of a is k plus this is a b. The value of a is k and b is four plus this b squared. So, b is four squared and this whole equation is equals to zero.

[00:02:47]And now, further in the next step, we need more simplifications here and we get k minus four times of this is this is k squared plus four times of k plus this four squared becomes 16 and this whole equation is equals to zero.

[00:03:11]And now, in the next step, we divide this equation into the two cases.

[00:03:17]The first case is we write k minus four is equals to k minus four equals to zero and the second case is we write k squared plus four k plus 16 equals to zero.

[00:03:38]So, we have these two cases. We solve or we find the roots of both of these two cases. So, here we have from this we move minus four to the right-hand side and we get the value of k is equals to plus four.

[00:03:53]So, this is the value of K.

[00:03:56]And now further in the next step, you see that this is the quadratic equation.

[00:04:02]And we solve this problem for three methods. You know about that.

[00:04:06]The first method is factorization method, second one is completing square method, and third one is by applying the quadratic formula. So, here we use the quadratic formula to solve this quadratic equation.

[00:04:19]And for this, we write the coefficients of this equation, and its coefficients are A is equals to one, B is equals to four, and C is equals to 16.

[00:04:32]And we state the quadratic formula as K is equals to minus B plus minus square root of B square minus four times of AC divided by two times of A.

[00:04:49]So, this is the our quadratic formula.

[00:04:52]And here we substitute all the values of A, B, and C. And this will be written as This is minus B. The value of B is minus four plus minus square root of B square. B is four square minus four times of A is one, and C is 16, and it is divided by two times of A, and here our A is equals to one.

[00:05:20]And now further in the next step, we have K is equals to minus four plus minus square root of four square. And four square becomes 16 minus four times of 16, and it is divided by two times of one is equals to two.

[00:05:42]So, now in the next step, we have K is equals to this is minus four plus minus Here we need to simplify the square root sign and inside the square root sign we have 16 be the common term. And when we take common 16, we get the remaining values are 1 minus 4 and it is divided by 2.

[00:06:05]So further in the next step here we need more simplifications and this will be written as Here we apply the square root on both of these two values and we get square root of 16 becomes four times of square root of 1 minus 4 and 1 minus 4 becomes minus three and it is divided by two.

[00:06:31]So further in the next step, we break this fraction and we apply the uh we apply this two separately on both of these two values.

[00:06:43]And this will be written as minus 4 divided by 2 plus minus four times of square root of three and we know that this uh when negative sign appears inside the square root sign, this will becomes iota and it is divided by two.

[00:07:03]So further we have this is these values are canceled out by each other and we get this is two times two becomes four.

[00:07:13]So here we have this is minus two plus minus this is two times two becomes four and we get two times of square root of three into iota.

[00:07:26]So here we get these are the two complex roots of the given equation.

[00:07:32]So here finally we have there are uh three roots of the given equation in which uh two roots are complex roots and uh, the one root are real root.

[00:07:49]So, the final three roots are K1 is equals to -2 + 2 * of square root of 3 into iota.

[00:08:01]K2 is equals to -2 -2 * of square root of 3 into iota and K3 is equals to 4.

[00:08:12]So, these are the three roots of the given equation and this is the final answer. And thank you so much for watching this video. Please subscribe to my channel for more exciting videos.