Video: CBSE Class X • Pack 2 • 2017 • Question 11

CBSE Class X • Pack 2 • 2017 • Question 11

08:52

Video Transcript

If 𝑎𝑑 is not equal to 𝑏𝑐, prove that the equation 𝑎 squared plus 𝑏 squared 𝑥 squared plus two times 𝑎𝑐 plus 𝑏𝑑 𝑥 plus 𝑐 squared plus 𝑑 squared equals zero has no real roots.

We have a quadratic equation. And we’d like to show that it has no real roots. The number of real roots that a quadratic equation has depends on the discriminant of that equation. The discriminant, which we’ll call 𝐷, of the general quadratic equation 𝑎𝑥 squared plus 𝑏𝑥 plus 𝑐 equals zero is 𝑏 squared minus four 𝑎𝑐.

If the discriminant 𝐷 is positive, then the equation has two real roots. If it is zero, then the equation has one real root. And if it is negative, then the equation has zero real roots.

We’re interested in showing that our equation has no real roots. And we can do this by showing that the discriminant is less than zero. But to show that the discriminant is less than zero, first we have to find the discriminant.

We know what the discriminant of 𝑎𝑥 squared plus 𝑏𝑥 plus 𝑐 is. It’s 𝑏 squared minus four 𝑎𝑐. Perhaps, we should change the letters that we’re using to stand for the coefficient of 𝑥 squared 𝑥 and the constant term as 𝑎, 𝑏, and 𝑐 already have meanings in the question. Changing 𝑎, 𝑏, and 𝑐 to 𝑙, 𝑚, and 𝑛 then, the discriminant becomes 𝑚 squared minus four 𝑙𝑛.

What is 𝑚? Well, it’s the coefficient of 𝑥 in the equation. And we can see in our equation that the coefficient of 𝑥 is two times 𝑎𝑐 plus 𝑏𝑑. We substitute this value for 𝑚. So 𝑚 squared becomes two times 𝑎𝑐 plus 𝑏𝑑 all squared.

We do the same for 𝑙. 𝑙 is the coefficient of 𝑥 squared, which is 𝑎 squared plus 𝑏 squared we can see. We make this substitution. And finally, 𝑛 is the constant term of our equation, which we can see is 𝑐 squared plus 𝑑 squared. Making this substitution then, we have the discriminant of our equation in terms of the unspecified numbers 𝑎, 𝑏, 𝑐, and 𝑑.

Now, remember we want to show that this discriminant is always less than zero no matter what the values of 𝑎, 𝑏, 𝑐, and 𝑑 are and hence that the equation has no real roots. That’s what the question asks us to do. Okay, so let’s clear some room so we can do this.

We have an expression for the discriminant capital 𝐷 in terms of the lowercase 𝑎, 𝑏, 𝑐, and 𝑑. But we can simplify this expression. For example, two times 𝑎𝑐 plus 𝑏𝑑 all squared is two squared times 𝑎𝑐 plus 𝑏𝑑 squared. And of course, two squared is just four. So the first term becomes four times 𝑎𝑐 plus 𝑏𝑑 squared.

We see now that the two terms have a common factor of four, which we can pull out. And while we do this, we can take the opportunity to write 𝑎𝑐 plus 𝑏𝑑 squared as 𝑎𝑐 plus 𝑏𝑑 times 𝑎𝑐 plus 𝑏𝑑. Now, we have some brackets so we can expand.

Starting with the first pair, we have 𝑎𝑐 times 𝑎𝑐 or 𝑎𝑐 squared plus 𝑎𝑐 times 𝑏𝑑 plus 𝑏𝑑 times 𝑎𝑐 plus 𝑏𝑑 times 𝑏𝑑 or 𝑏𝑑 squared. We expand the other pair of brackets similarly. We have some redundant brackets here. And we can get rid of the other brackets by distributing the minus sign over them. All the plus signs become minus signs.

We can write some of our terms more conventionally. For example, we can write 𝑎𝑐 all squared as 𝑎 squared times 𝑐 squared. Similarly, by convention, rewrite the variables in a term in alphabetical order. So 𝑎𝑐𝑏𝑑 becomes 𝑎𝑏𝑐𝑑. The term 𝑏𝑑𝑎𝑐 also when written in alphabetical order is 𝑎𝑏𝑐𝑑.

Writing terms in this conventional way helps us to spot like terms which we can combine. We write 𝑏𝑑 all squared as 𝑏 squared times 𝑑 squared. The other terms are already written conventionally and so don’t require any changes.

Apart from the like terms that we’ve already mentioned, there’s some other simplification we can do. An 𝑎 squared 𝑐 squared cancels with a minus 𝑎 squared 𝑐 squared and 𝑏 squared 𝑑 squared cancels with a minus 𝑏 squared 𝑑 squared. Combining the like terms then and tidying up inside the brackets, we get two 𝑎𝑏𝑐𝑑 minus 𝑎 squared 𝑑 squared minus 𝑏 squared 𝑐 squared.

Now, it may not be so obvious how to continue. Remember we want to show that this discriminant is negative. That is it is negative for all values of 𝑎, 𝑏, 𝑐, and 𝑑. The trick is that for any real number 𝑛, 𝑛 squared is greater than or equal to zero. And we can change this to the strict inequality 𝑛 squared is greater than zero if we exclude the case that 𝑛 is equal to zero.

We need to show that the discriminant is less than zero. And so we modify our factor slightly. We can say that negative 𝑛 squared is less than zero for any nonzero real number 𝑛. And of course, we can replace negative 𝑛 squared by negative two 𝑛 squared or negative three 𝑛 squared and so on.

Now, is it obvious that we have something of this form here. Let’s pull out a factor of negative one. We can also reorder the terms inside the brackets. And it may not be obvious, but it is true, that what we have inside the brackets is a perfect square.

There’s a hint in the question. We’re told that 𝑎𝑑 is not equal to 𝑏𝑐. And we also know that our square number trick only works if the real number we’re squaring is not zero. If we write 𝑎 squared 𝑑 squared as 𝑎𝑑 all squared, 𝑎𝑏𝑐𝑑 as 𝑎𝑑 times 𝑏𝑐, and 𝑏 squared 𝑐 squared as 𝑏𝑐 all squared, then we might be able to see more easily that we have a perfect square.

Now, we know that 𝑥 minus 𝑦 all squared is equal to 𝑥 squared minus two 𝑥𝑦 plus 𝑦 squared for all values of 𝑥 and 𝑦. And we can recognise what we have inside the brackets at the right-hand side of this identity with 𝑥 equal to 𝑎𝑑 and 𝑦 equal to 𝑏𝑐. And so we can substitute 𝑎𝑑 for 𝑥 and 𝑏𝑐 for 𝑦 in the left-hand side of our identity to get that our discriminants capital 𝐷 is negative four times 𝑎𝑑 minus 𝑏𝑐 squared.

Now that we’ve written capital 𝐷 the discriminant in this way, it’s straightforward to see that it is negative, at least that is when 𝑎𝑑 minus 𝑏𝑐 is nonzero. If 𝑎𝑑 minus 𝑏𝑐 is zero, then the discriminant 𝐷 is also zero. But the condition that 𝑎𝑑 minus 𝑏𝑐 is nonzero is the same as the condition 𝑎𝑑 is not equal to 𝑏𝑐. And we’re told that 𝑎𝑑 is not equal to 𝑏𝑐 in the question.

This is from a generalisation of our fact that negative 𝑛 squared is less than zero if 𝑛 is not equal to zero. Perhaps, we should make this explicit: if 𝑠 is any negative number, then 𝑠 times 𝑛 squared is less than zero if 𝑛 is not equal to zero.

Now, why did we want to show that the discriminant 𝐷 is negative? It’s because the sign of the discriminant tells us how many real root a quadratic equation has. If the discriminant is positive, the equation has two real roots. If the discriminant is zero, then it has one real root. And if as in this case the discriminant is negative, then the equation has no real roots. And this is what we were required to prove.

Let’s recap then. We found the discriminant capital 𝐷 of our quadratic equation in terms of the arbitrary numbers 𝑎, 𝑏, 𝑐, and 𝑑. Straightforward simplification got us to this line. But some ingenuity was required to write this as negative four times 𝑎𝑑 minus 𝑏𝑐 squared.

But having written the discriminant in this form, it was straightforward to see that the discriminant is negative when 𝑎𝑑 is not equal to 𝑏𝑐 as we’re told it is the question. And as its discriminant is negative, we know that our quadratic equation has no real roots.

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