If α, βare the distinct real roots of a quadratic equation a x^2+b x+c=0 where a, (c)/(2), b are

If α, βare the distinct real roots of a quadratic equation a...

Key Concepts

Quadratic Equations and Vieta's Formulas

A quadratic equation in the form ax² + bx + c = 0 has roots that can be connected through Vieta's formulas. These formulas relate the sum and product of the roots to the coefficients of the equation, which is essential for understanding the deeper relationships between the coefficients and the behavior of functions or sequences defined by those roots.

Recurrence Relations for Power Sums

When dealing with sequences defined by terms like t? = ?? + ?? (where ? and ? are roots of a quadratic), the sequence satisfies a linear recurrence relation. This recurrence is derived from the fact that the powers of the roots obey the initial quadratic's property, allowing one to express higher powers as combinations of previous ones. This concept is fundamental in studying sequences in algebra and number theory.

Homogeneous Linear Systems

A homogeneous system of linear equations is one in which all equations are set equal to zero. Such systems always have at least the trivial solution (where all variables are zero). Whether non-trivial solutions exist depends on the relationships among the coefficients, often determined by the rank or determinant of the coefficient matrix.

Determinants and Consistency Conditions

The determinant of the coefficient matrix of a system of linear equations is a key factor in deciding the nature of its solutions. A nonzero determinant indicates that the system has only the trivial solution, while a zero determinant typically points toward infinitely many solutions. Evaluating the determinant is thus essential for understanding whether a system of equations is consistent and if non-trivial solutions exist.

Transcript

00:01 So in the given question we are told that alpha and beta are the distinct real roots of the quadratic equation.

00:09 A x squared minus bx plus c is equal to 0.

00:16 And we are told that the numbers a, c by 2 and b are not in an ap.

00:28 And we are also told that if there is a relation such that tk is equal to alpha raised to k plus beta raised to k and where k is greater than or equal to 1 we have to find what kind of solutions does the system of equations that are given as 3x plus 1 plus t1y plus 1 plus t2 z equal to 0, 1 plus t2 z equal to 0 is the first equation of the system of equations.

01:08 And we have the second equation as 1 plus t1 x plus 1 plus t2y, y plus 1 plus t3 z, 1 plus t3 z is equal to 0 and we have the last equation of the system which is given as 1 plus t2x plus 1 plus t3y y plus 1 plus t 4 z is equal to 0 1 plus t 4 z is equal to 0 so we have to find what kind of solutions does this system of equations have so what we are going to start with is we know that alpha and beta are the distinct real roots of the quadratic equation a x squared minus b x plus c equal to 0.

02:09 So if that's the case when we take since they are distinct real roots the condition for a quadratic equation to have distinct real roots is b squared minus 4a c equal to 0.

02:22 So while we take the solution what you will have is minus b plus or minus square root of b square minus 4ac c divided by 2a and since we have distinct real roots we can write the value of b square minus 4ac as as 0 so the solutions would be minus b divided by 2a.

02:46 So over here we can see that the the coefficient of x.

02:53 Over here that is denoted over here as b is minus b right so minus of minus b divided by 2a let this be alpha and we will have beta also equal to b by 2a right minus of minus b is b so b by 2a so now when we add alpha and beta we can find a relation that alpha plus beta is equal to plus b by a and similarly we can write alpha times beta is equal to c by a right.

03:32 So these are the two relations that we can write knowing that alpha beta are the distinct real roots of the given equation given quadratic equation right and now let's move on to the system of equations that we have so let's keep these relations in mind right these two relations and the system in the system of equations if the system of the form a x equal to zero has a trivial solution as has a trivial solution then the condition for that is that the determinant of the determinant of a would give us a value that is not equal to 0 right? so now let's take the determinant of the matrix that we can form from the coefficient of the coefficients of the variables in the system of equation.

04:39 So we can look at the equations and we can take the coefficients one by one.

04:45 So in the first equation the coefficient are coefficients are 3, 1 plus t 1 and 1 plus t 2.

04:53 Next we have 1 plus t 1 plus t 2.

04:54 Next we have 1 plus t 1 plus t 2, 1 plus t 3, and then we have 1 plus t 2, 1 plus t 3 and 1 plus t 4.

05:08 So this is the required determinant a, right? so now let's substitute for t 1, t 2, t 3 and t 4.

05:20 Since we have the relation that t k is equal to alpha raised to k plus b.

05:26 Beta raised to k.

05:28 So then we will have the determinant 3.

05:31 1 plus alpha plus beta 1 plus alpha plus beta square plus then over here is 1 plus alpha plus beta plus 1 plus alpha square plus beta square and 1 plus alpha cube plus beta cube.

05:51 Next 1 plus alpha square plus beta square.

05:55 1 plus alpha cube plus beta cube 1 plus alpha raised to 4 plus beta raised to 4.

06:05 So this is what we get after the substitution of t1, t2, t3 and t4.

06:14 So now what we are we are going to do is we can split this determinant into two parts, right? so, we can split it as we can split this determinant and write it as first let's write the determinant as 1 plus 1 plus 1.

06:37 So what we are going to do is we can take this determinant which is over here let's take this determinant and before splitting the determinant let's write the number 3 we can write the number 3 over here as 1 plus 1 plus 1.

07:08 Right? so we can write the number 3 as 1 plus 1 plus 1 plus 1 and what we are going to do next is we are going to split this determinant as we can write this as 1 1 1 1 1 1.

07:36 111 111 which gives us we can write we can split this determinant and write it as 1 1 1 1 alpha beta 1 alpha b2 times 1 1 alpha 1 alpha 1 alpha square beta times 1 1 alpha 1 alpha alpha square and 1 beta beta square so how did we split this so what we have over here is two determinants and when we take the multiplication of this determinant what we will have is we are going to take the multiplication as we are going to take the first element 1 and multiply it with the first element of the second matrix that is 1 so we are we will have 1 over here plus the second element and the second element of the first column in the second matrix...

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