Video: CBSE Class X • Pack 4 • 2015 • Question 7

CBSE Class X • Pack 4 • 2015 • Question 7

04:32

Video Transcript

Solve the following quadratic equation for 𝑥: four 𝑥 squared plus four 𝑏𝑥 minus 𝑎 squared minus 𝑏 squared is equal to zero.

Now, as we don’t know the values of 𝑎 and 𝑏, our solution for the roots of this quadratic equation will be in terms of 𝑎 and 𝑏. We have a variety of methods that we can use to solve a quadratic equation. We could try factorizing, we could use the quadratic formula, or we could complete the square. Now, this quadratic equation doesn’t look like it will easily factorize with simple algebraic expressions. So instead, we’ll try applying the quadratic formula.

The quadratic formula tells us that the roots of the quadratic equation, 𝑎𝑥 squared plus 𝑏𝑥 plus 𝑐 is equal to zero, are given by 𝑥 is equal to negative 𝑏 plus or minus the square root of 𝑏 squared minus four 𝑎𝑐 all over two 𝑎. Now, we need to be a little bit careful here because the letters 𝑎 and 𝑏 appear in the quadratic formula, but they also appear in our quadratic equation. And they have different meanings in the two.

For our quadratic equation, the value of 𝑎, which is the coefficient of 𝑥 squared, is four. The value of 𝑏, which is the coefficient of 𝑥, is four 𝑏. And the value of 𝑐, which is the constant term, is negative 𝑎 squared minus 𝑏 squared. Now, we can actually simplify the expression for 𝑐 by multiplying every term in the bracket by this negative sign or negative one. And it gives 𝑏 squared minus 𝑎 squared.

So we can substitute our values of 𝑎, 𝑏, and 𝑐 into the quadratic formula. And it gives 𝑥 is equal to negative four 𝑏 plus or minus the square root of four 𝑏 all squared minus four multiplied by four multiplied by 𝑏 squared minus 𝑎 squared all over two multiplied by four. We now just need to work through simplifying this expression for the two roots of the quadratic equation, carefully.

Let’s look at simplifying within the square root first of all. The first term within the square root is four 𝑏 all squared. Now this is equal to four squared multiplied by 𝑏 squared which is 16𝑏 squared, not four 𝑏 squared. You must make sure you square the four as well as the 𝑏. So watch out for this common mistake here.

The second term within the square root is four multiplied by four multiplied by 𝑏 squared minus 𝑎 squared. Well, four multiplied by four is 16. And then we need to multiply this by each term inside the bracket. So it gives 16𝑏 squared minus 16𝑎 squared. So our expression for 𝑥 becomes negative four 𝑏 plus or minus the square root of 16𝑏 squared minus 16𝑏 squared minus 16𝑎 squared all over eight.

Now, within the square root we can multiply both terms inside that bracket by the negative or negative one sign in front of it. And it will give 16𝑏 squared minus 16𝑏 squared plus 16𝑎 squared. The two lots of 16𝑏 squared, one of which is positive and one of which is negative, cancel each other out. So within the square root, we’re just left with 16𝑎 squared. This means that our expression for 𝑥 simplifies to negative four 𝑏 plus or minus the square root of 16𝑎 squared all over eight.

Now to find the square root of 16𝑎 squared, this is equal to the square root of 16 multiplied by the square root of 𝑎 squared. And the square root of 16 is four. And the square root of 𝑎 squared is 𝑎. So the square root of 16𝑎 squared is just four 𝑎. Our expression for the roots, therefore, becomes 𝑥 equals negative four 𝑏 plus or minus four 𝑎 all over eight. And it’s starting to look a lot more simple.

We can also cancel a common factor of four from every term. So we’re left with negative one 𝑏 or negative 𝑏 plus or minus one 𝑎 or 𝑎 in the numerator and two in the denominator. These roots are now fully simplified. So the solution to the given quadratic equation is 𝑥 is equal to negative 𝑏 plus 𝑎 all over two or negative 𝑏 minus 𝑎 all over two.

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