Alright, solve it and get the alpha value. And obviously, 1 alpha value will be rejected. If you complete it yourself, then you will say, ma'am, how will we reject alpha value? 3 alpha root should be there, right? So that means, 3 alpha which you will get, it should be satisfied.
Now for one of the values, it won't be satisfied. That means, that value of alpha will be rejected. Alright, so this is the answer.
minus 3 by 2 minus 2 by 4. Now, there is a problem. Which is the root of 3 is to 4? This is nothing but minus 4 by 2. So, these are the two roots that are in the ratio 3 is to 4. Alright.
So, next is an interesting one. So, x raised to the power of 4 minus 10x square plus 9x minus 2 equal to 0 given that the product of its 2 roots given that the product of its 2 roots is unity. If you use direct method, see what happens to you.
3 power 4 minus 5 minus 9x minus 2 equals 0. And product of its two roots is unity. So if I take one root to be alpha, the other root will be 1 by alpha. By quadratic here. So that means I need to take theta as well as gamma. Right?
Now, as soon as I see relationships, I will have 4 relationships. 1, 2, 3, 4. Now you will say, ma'am, 4 equations, 3 variables, we can easily solve it. Really?
Easily? By solving it easily, won't we be able to make a cubic equation? In fact, we will make a cubic equation. Now you have changed the word by quadratic state to cubic.
Still, you will have to apply methods. Which cubic can we solve? If you apply the method of two back to square one in the different trials, then we will have to think of something else.
What else to think? Now, is there any benefit of always using relationships? That depends on the question. Now, there is a bi-political equation. Product of its two roots is unity.
What is it being used for? Use the y-quadratic equation. If I know its two roots, obviously it will become quadratic.
I can easily solve it. But again, the problem is how to make it quadratic. I should know its two roots. Okay. You are absolutely right.
How? Now, tell me one thing. This is a y-quadratic polynomial.
Right. It will have four linear factors and two quadratic factors. There will be two quadratic factors and you know what is it?
The product of its two roots is unity. So if I write the quadratic factor from those two roots, I will know the kind of form. Imagine, now what I want to do is, I want to write this biquartic polynomial as a product of two quadratic factors.
How will they fill up there? You are absolutely right by using common sense. Look, first of all, I have to write a quadratic factor.
That means x squared, x squared. Deeding coefficient is 1. So, I am keeping the deeding coefficient 1 here. Right. Now, you will say, ma'am, two turns, two turns. How?
You are absolutely right. Product of two of its roots is unity. If I assume that two roots come from here.
Product of its root is unity. That means, what should be the constant term here? Plus 1. Product is plus 1. Right? Now, if I tell you one thing, here a plus is made.
Constant term will be multiplied by the constant term here. That's why I will get this constant term. I have to make a constant term. Constant term and constant term will be multiplied. So, tell me, how will I multiply 1 by minus 200?
Answer is very simple. Minus 200. Now, you will say, ma'am, that's all right. Now, you don't have to tell the coefficients of the x-value term. Absolutely right.
Let me tell you, let me assume. Let's assume this is Ax and this is B. I have to get the values of A and B.
Let's do it. Now, let's get the values of A and B. What have I done?
I have resolved this polynomial into two quadratic factors. So, I am multiplying them. Do you want to multiply?
Excuse me. What is the function of the... Now we have written cubic.
Now we have come to the 10th class so quickly we will write cubic. Now from where will the cubic terms come? From where do I multiply x square?
From dx. So it will become cubic. So that means what will be the coefficient?
b. With whom do I multiply x? That I will get cubic.
From x square. So what will be the coefficient here? a. So as in b plus a into x2 plus. Now x square.
Now, with whom should I multiply the x square here? With minus 2. Right. Will I get more coins? Yes. With whom should I multiply ax?
With dx. So, plus ab. More coins?
So, if I multiply 1 with x square, then also I get x square. So, plus 1. Right. Now, let's write the term of x. Right.
So, x. So, with whom should I multiply x square? With x. Now, this polynomial, this entire polynomial is product of these two factors. So, what is the polynomial of product of factors?
minus 10x2 plus 9x minus 2. Two polynomials are equal. Why don't I use that? Because two polynomials are equal. So that means corresponding coefficients are equal.
Now 1 is equal to 1. Here b plus a will not be equal. It is an x cube term. But x cube's coefficient will be 0. 0 into x cube.
So that means b plus a. should be equal to 0 right. Now, x square of coefficient x square of coefficient these two should be equal as in minus 2 plus ab plus minus 2 plus ab plus 1 minus 1 directly minus 1 yeah ab minus 1 equal to minus n right and then this is equal to this rather than minus 2 a plus b equal to. and minus 2 minus 2 both the same thing. So, I have 3 equations.
2 variables. Now, tell me which 2 equations to solve easily. So, what is the quality? Now, this is linear and this is linear. Definitely, I will do this.
So, I will write it like this. b minus 2a equal to n. Right. Now, what should I do? Subtract this.
So, sine change. Sine change. Done. change.
From here I will get a value. a will turn out to be minus 3 and consequently b will turn out to be 3. So minus 3 and b is 3. That's it. Now I have to solve this.
Solve y quadratic for solve. Think about it. Now you have resolved your problem into this either this is 0 solution x square minus 3x plus 1 equal to 0 solution you can easily solve it or x square plus 3x minus 2 equal to 0 solution.
So, what is our solution now you have to decide whether this method is better or getting stuck in the cubic equation whether that is better. This one and this one is more fun as well So, I am thinking what should I keep here and what should I keep here. I don't want to sing the values of a and b. I think this is more fun. I mean for me.
I enjoy this mission better. Alright, try on the next one. Solve x raise power 4 plus 12x minus 5 equal to 0. Now you will say ma'am, you don't know the condition, tell us.
Now what to do? The method is the same. I need to write this polynomial as a product of two coordinate polynomials. Let's think. Now obviously, x square will come here and x square will come here.
Right? x is equal to 4 and leading coefficient is 1. Now it is possible that we knew something here, so we guessed the constant. Now nothing is known here.
Pid ek baah batao, yaha pe hai minus 5. Right? Abhi tak jo main quality click se aare ho, uski coefficient se integral. Right?
So minus 5 ko, I need to write it as product of two integers. Keshis kya ho sakta? Minus 5. Yaha to mai likhsakte 1 into minus 5. Or I can write it as minus 1 into 5. Keshis hai?
Minus 5 ko, yaha to mai likhsakte 1 into minus 5. Yaha minus 5. minus 1 into 5. Right? These are the only two possibilities. There are two possibilities. I will consider both of them.
How much time will it take? So, first case, I assume that 1 into minus 5 is the appropriate one. So, here I will take 1 minus 1 into minus 5. Now, you will say, ma'am, again, will you assume the x-value term?
Absolutely, I will assume. So, I will suppose this to be ax and this to be bx. Right. Multiply.
So, comparing the coefficients, we will be able to ascertain something. Right. Now, multiply.
Multiply. Multiply. Multiply.
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Multiply. Multiply. Multiply. Multiply. Multiply.
Multiply. Multiply. Multiply. Multiply. Multiply.
Multiply. Multiply. Multiply. Multiply.
Multiply. Multiply. Multiply. Multiply. Multiply.
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Multiply. Multiply. Multiply.
Multiply. So this is minus 5 plus a b plus 1. So this is written as minus 4 into x square plus 1. Don't write x as a term. So minus 5 a plus b into x. Now this is minus 5. Obviously. Now this polynomial with whom you have to sit?
With this. So as in x raised to 4. which is equal to x raise power 4 plus 12x minus 5 obviously 1 1 again excuse me because 12th elliptic is 0 into x plus 0 into x right. So this is equal to this as in a plus b equal to 0 and minus 4 plus a b equal to 0 and this is equal to 12 minus 5 a. So, minus 5 a plus b equal to 12 and minus 5 minus 5 right.
Now, though three equations are two variables and equations are two equations, once we consider them, two linear equations are considered. So, minus 5 a plus b equal to 12. Now, subtract correct, times change, times change, times change, a will turn out to be minus 6. minus 2. Right? a will turn out to be minus 2. So consequently, b, I will put this equation here, b will turn out to be 2. And I will tell you that this is the solution.
You will have to check. Why? Why did I solve this? Why did I solve this equation?
I solved these two equations. I want a and b, which satisfy this equation. This equation as well as this equation, this equation should also be solved.
This equation should be satisfied. Now let us see whether this equation is being satisfied or not. So this is minus 4 plus, a is minus 2 into d is 2. Now who got this?
Minus 4 minus 4 has been minus 8. Minus 8 is definitely not equal to 0. So that means the values of A and B are not satisfied in the third equation. What was the moral of the equation? That means, we assumed that this case will not happen.
Now there should be another case. Now you will say, ma'am we have worked so hard. It's not real. Sometimes it works, sometimes you have to work harder. There is a big deal that there is a big deal that there is a big deal.
So, this becomes x square plus dx minus 1 into x square plus dx plus 5. Now, let us multiply. By the way, I have taken a big cost. To multiply, there will be 6 terms but I have to change it a little. So, I have taken x square plus dx minus 1 into x square plus dx plus 5. Now, let us multiply.
By the way, I have taken a big cost. To multiply, there will be 6 terms but I have to change it a little. So, let us multiply.
So, let us multiply. So, let us multiply. So, let us multiply. So, let us multiply.
So, let us multiply. So, let us multiply. So, let us multiply. So, let us multiply.
Let's see. Let's take the multiplier. Let's not. Otherwise chances will increase. Right.
So now we are multiplying them. This case is done. So this becomes x to the power 4 plus x cube term. Obviously, c plus a. And a plus b into x cube.
Then x. Square term. So, 5 plus AB minus 1. Yeah.
4 plus AB into X plus. Plus. X term.
5 X minus B. Minus 5. Now, with whom does this compare? With this polynomial.
X is for 4 plus 12 X minus 5. 1 1 this should be a plus b equal to 0, then 4 plus a b equal to 0 and 5 a minus b equal to 12. Obviously, in over question 5 a minus b equal to 12 add the 2. So, this becomes 6 a equal to 12. So, a becomes equal to 2 and b is minus 2. Now, I have solved both these and got the values of a and b. Has it satisfied the technical dos? So, 4 plus a is not required.
4 plus a is 2. So, 2 into b becomes 2 into minus 2. That is minus 4. So, this is definitely equal to 2. It will satisfy. So, that means this will be my value of a. And now, Now, equate this to 0 and find the answer.
There are two quadratic factors. Easily you can find the answer. Do you need to write the answer? Yes. See, its discriminant is less than 0 or greater than 0. You will not get the real roots from here.
Just solve it and you can easily solve it. So, write down the next one the last question for write down come on last question. of p and q over tallies of p and q the polynomial x raised to the power minus 14 x cube plus 71 x square 71 x square plus p x e x plus q is the perfect square. Any comments on this question? Come on.
Alright. So he says for what values of P and Q does this polynomial become the perfect square? Polynomial becomes the perfect square.
Actually, this is a bi-qualitative polynomial. To what square should I do? To what degree of polynomial should I do a square to get a bi-qualitative polynomial?
Obviously, I will do a quadratic polynomial. And here, x raised to the power 4, I will take x raised to the power 2. As in, I will do x squared to the power 4. Right? Now, Quadratic polynomial ka square khanna. Clearly, mujhe nahi pata ki yahaan pe linear term and constant term ke aane wali. Koi baat nahi.
Suppose, this is AX, this is B. Suppose that this polynomial is equal to X square plus AX plus B whole square. Achhar, itse whole square expand dana toh aata hai na. Yahaan se jo polynomial hai aur ye wala jo polynomial hai. Same hai.
Again. You can compare the coefficients. So, first of all, expand this.
So, to expand this, a plus b plus c whole square, which is equal to a square, as in x square's square, as in x square's whole, plus b square. b square is a square x square, plus a square x square plus b square, plus 2 into this into this, as in... 2 into 2ab plus 2b. This into this. 2abx plus 2bx squared.
Expand this. Now this polynomial is definitely equal to this one. Before writing this, let's rearrange it.
This is the force of x-ray. Plus 2axq plus a squared. plus 2b into x square plus 2abx plus b square.
So, that comparison is easier. Now, this is equal to this polynomial. This is equal to x is equal to 4 minus 14x cube plus 71x square plus 3x plus q.
So, two polynomials are equal. 1 month is the secret. And now, Now 2a should be equal to minus 14. 2a equal to minus 14. So a value again minus 7. No.
Now next. This is a square plus 2b equal to 71. a square plus 2b equal to 71. So a value is also known. So a is minus 7. a square is 49. So 2b equal to 22. Right.
So, b will turn out to be 11. So, a b is known and b b is known. Now, t is nothing but q a b equal to t. We had to find the value of t, right? I knew it from the start.
So, this is my t is equal to minus 1 to t square. And you do this. And q is equal to t square or d square equal to q.
Now, b is 11. So, q is equal to 121. For these values of P and Q, this polynomial is the square of this polynomial. The n is also there. Insert the number. So, it is an easy question.
We are ending it on an easy question. As you can see, today's lexicon was very easy. It was fun for me.
I enjoyed doing this a lot. Right, so in today's lecture, what we did is, we know a quadratic equation to solve. In today's lecture, we learned, not learned, we looked at some methods to solve a cubic or a biquadratic equation.
Right, polynomial equation. Right, what methods can be used for that? Either there will be integral roots, so we can find that out directly.
Or there will be rational roots, we can directly find it out. And then next. between coefficients and zeroes, roots right. And the next the most fun part we can write the quadratic polynomial as a product of two, no we can write a bi-quadratic polynomial as a product of two quadratic polynomials. This means yes, we have to do it and yes, possibilities exist right.
So, that is it I hope you liked today's lecture. I hope you. You like the Fermat's last theorem part. That is something I generally like.
I enjoyed it. So that is of course homework. Homework is home assignment free.
So that's it and I will see you next time. Thank you.