Video: Solving Quadratic Equations That Include Radical Roots

Solve 𝑥² + 3𝑥 − 6 = 0.

02:39

Video Transcript

Solve 𝑥 squared plus three 𝑥 minus six equals zero. So exactly the same technique; it’s just that towards the end of this question the numbers are gonna get a little bit trickier. But never mind we’ll work out how to do that when we get there. So we’ll clear off a little space on the left-hand side by adding six to both sides and then we’ll complete the square on the left-hand side. So we’ve got 𝑥 and we take half of the 𝑥 coefficient, three over two; we’re squaring that. To make that equivalent to the expression above it, we’re gonna need to take away that three over two term all squared. So that’s equal to six. So I’m just gonna evaluate the three over two all squared. So I’ve got 𝑥 plus three over two squared minus nine over four is equal to six and I’m gonna add the nine over four to both sides. So I’ve got six plus nine over four on the right-hand side. So I need to find an equivalent version of six, which is actually a top heavy fraction over four. So I’ve got my common denominators. And that’s obviously twenty-four over four because twenty-four over four is the same as six. Now twenty-four over four plus nine over four equals thirty-three over four. So I’ve got 𝑥 plus three over two squared is equal to thirty-three over four. Now I’m gonna take square roots of both sides. And the square root of the left-hand side is just 𝑥 plus three over two and we got two possible values for the right-hand side: it’s either the positive version of the square root of thirty-three over four or it’s the negative version. Now obviously that’s the same as root thirty-three over root four. So that’s the root of four is two.

So this is slightly different to the previous questions because always before when we were doing this square rooting stage, we ended up with nice simple numbers. But the square root of thirty-three isn’t anything nice and easy. So we just leave it as root thirty-three. And in the next step to get 𝑥 on its own, on the left-hand side here I’m subtracting three over two from both sides. And so the right hand becomes negative three over two plus or minus root thirty-three over two. So as luck would have it here, I’ve already got my common denominator. So I’ve now got two possible values for 𝑥: it could be negative three over two plus the root of thirty-three over two or it could be negative three over two subtract this root of thirty-three over two. Now I could gather those terms together but because these numbers are not particularly nice, I have to leave them in-in that sort of format. Now the question might-might have asked me to give my answers rounded correct to one decimal place or it might have asked me to leave them in this exact format. And just for the sake of argument, I’m gonna pretend they asked me to leave it in the exact format in this case. So sometimes the answers work out really simply and sometimes the numbers are a little bit more complicated, but the method is just the same.

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