# Video: CBSE Class X • Pack 4 • 2015 • Question 12

CBSE Class X • Pack 4 • 2015 • Question 12

05:43

### Video Transcript

Solve for 𝑥: root three 𝑥 squared minus two root two 𝑥 minus two root three is equal to zero.

Now, this equation is a quadratic equation as the highest power of 𝑥 is a two. We have this 𝑥 squared term here. We have a choice of methods for how we can solve a quadratic equation: we could try to factorise it, we could use the quadratic formula, or we could complete the square.

In this quadratic equation, the coefficients are not particularly straightforward. They’re given in terms of surds, which means it’s probably not going to factorize easily. Our best option then is to apply the quadratic formula. The quadratic formula tells us that the roots of the general 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.

Here, 𝑎 is the coefficient of 𝑥 squared, 𝑏 is the coefficient of 𝑥, and 𝑐 is the constant term. So for this quadratic equation, 𝑎 is equal to root three, 𝑏 is equal to negative two root two, and 𝑐 is equal to negative two root three. So we can substitute the values of 𝑎, 𝑏, and 𝑐 into our quadratic formula. And it gives two root two first of all because this is negative negative two root two plus or minus the square root of negative two root two squared — that’s 𝑏 squared — minus four multiplied by root three multiplied by negative two root three — that’s four 𝑎𝑐 — all over two 𝑎 which is two root three.

Now, we need to simplify this. And we can start underneath the square root first of all, where we have negative two root two squared. Now, this is equal to negative two root two multiplied by negative two root two. And we can do these multiplications in any order. So if we multiply negative two by negative two first, this gives four and root two multiplied by root two gives two. So we have four multiplied by two which is equal to eight.

Also within the square root, we have that bracket which is four multiplied by root three multiplied by negative two root three. So using the same idea, we can multiply the four by the negative two, which gives negative eight and then root three by root three which gives three. So we have negative eight multiplied by three which is negative 24.

Within the square root then, we have eight minus negative 24 and those two negative signs together make a positive. So we have eight plus 24, which is equal to 32. So the value within that square root is actually simplified very nicely. And we’re left with 𝑥 is equal to two root two plus or minus root 32 all over two root three.

Next, we want to see if we can simplify this root 32 any further. And to simplify a surd or a square root, we want to look for the square factors of the number underneath the square root. 32 is of course equal to 16 multiplied by two. And 16 is a square number. So we can express the square root of 32 as the square root of 16 multiplied by the square root of two.

Now, remember 16 is a square number. It’s equal to four squared. So the square root of 16 is just four. And therefore, root 32 simplifies to four root two. So the value of our root simplifies to two root two plus or minus four root two all over two root three. Now, we can actually cancel a factor of two from every term in this expression. And it leaves us with root two plus or minus two root two all over root three.

So our two unsimplified roots then, firstly, we have root two plus two root two all over root three and secondly, we have root two minus two root two all over root three. Now, the numerators of each of these roots can be simplified. For the first root, we had root two plus two root two which gives three root two and for the second root we had root two minus two root two which gives negative root two.

So our two roots become three root two over root three and negative root two over root three. Now, both of these roots currently have a surd in the denominator. And so we need to rationalize this. To rationalize, we need to multiply by root three over root three. Now, this doesn’t change the value of either root as this fraction is equivalent to one.

For the first root, we have three root two multiplied by root three in the numerator, which gives three root six, and root three multiplied by root three in the denominator which gives three. And now, we’ve rationalized the denominator. We can, however, simplify this root slightly further as the factor of three in the numerator can cancel with the factor of three in the denominator. And we’re just left with root six. This can’t be cancelled any further as six doesn’t have any square factors.

For the second root, we have negative root two multiplied by root three in the numerator, which is equal to negative root six, and in the denominator root three multiplied by root three which is equal to three. Now, this can’t be simplified any further. But it now has a rational denominator.

So our solution to the given quadratic equation is 𝑥 is equal to root six or negative root six over three.