Video: Evaluating a 2 Γ— 2 Determinant to Solve a Quadratic Equation

Solve the equation |9π‘₯Β², 2π‘₯ and 63π‘₯, 9π‘₯Β²| = βˆ’49.

04:22

Video Transcript

Solve the equation, the determinant of the matrix nine π‘₯ squared, two π‘₯, 63π‘₯, nine π‘₯ squared is equal to negative 49.

So, we can see in this question, we’ve got a two-by-two matrix. And we want to find the determinant of it. But we’ve got a general rule for finding the determinant of a two-by-two matrix. And that is, if we have the matrix π‘Ž, 𝑏, 𝑐, 𝑑, and we want to find the determinant of it, what we do is we multiply π‘Ž and 𝑑 and then subtract from it 𝑏 multiplied by 𝑐.

So therefore, if we want to find the determinant of our matrix, what we’re going to do is multiply the top-left term by the bottom-right term, so nine π‘₯ squared multiplied by nine π‘₯ squared. And that’s gonna give us 81π‘₯ to the power of four. That’s because if we have nine multiplied by nine, that’s 81. And then, π‘₯ squared multiplied by π‘₯ squared is π‘₯ to the power of four. And then, we’re gonna subtract from it two π‘₯ multiplied by 63π‘₯, which is gonna give us 126π‘₯ squared. And that’s because two multiplied by 63 is 126, π‘₯ multiplied by π‘₯ is π‘₯ squared.

So, now, what we’re gonna do is substitute the determinant of our matrix back into the original equation. And when we do that, we have 81π‘₯ to the power of four minus 126π‘₯ squared equals negative 49. So, what we’re gonna do now is add 49 to each side of the equation. And when we do that, we get 81π‘₯ to the power four minus 126π‘₯ squared plus 49 is equal to zero. So, we can see here that we got π‘₯ to the power four. But actually, in fact, there is a hidden quadratic within our equation here.

And to find it, what we can do is substitute in 𝑒 for π‘₯ squared. So then, our first term is gonna be 81𝑒 squared. And that’s because if 𝑒 is equal to π‘₯ squared, well, if you square π‘₯ squared, you get π‘₯ to the power four. And then, you have minus 126𝑒. And that’s because 𝑒 is equal to π‘₯ squared. Then, plus 49 is equal to zero. So, now, we formed a quadratic. So, what we can do is use the quadratic formula. And that formula is π‘₯ is equal to negative 𝑏 plus or minus the square root of 𝑏 squared minus four π‘Žπ‘ over two π‘Ž. And that’s when we have a quadratic in the form π‘Žπ‘₯ squared plus 𝑏π‘₯ plus 𝑐.

So therefore, in our equation, our π‘Ž is equal to 81, our 𝑏 is equal to negative 126, and our 𝑐 is equal to 49. So therefore, if we substitute in our values, we’re gonna have 𝑒 is equal to negative negative 126 plus or minus the square root of negative 126 all squared minus four multiplied by 81 multiplied by 49 all over two multiplied by 81. So, this is gonna be equal to 126. And that’s cause we had negative negative 126, and two negatives make a positive. Then, plus or minus the square root of zero over 162.

And then, what we can do is divide both the numerator and denominator by 18 cause 18 is a factor of 826 and 162. So therefore, we get 𝑒 is equal to seven over nine. So, great, we solved the equation. Well, no, because we’ve just found 𝑒, and we want to find π‘₯. Well, we know that 𝑒 equals π‘₯ squared cause that’s our substitution from earlier. So therefore, π‘₯ squared is equal to seven over nine, or seven-ninths. So therefore, if we take roots of both sides of equation, we’re gonna get π‘₯ is equal to plus or minus the square root of seven over nine, or square root of seven-ninths.

Well, we can now use one of our rules that we know. And that is, if we have root π‘Ž over 𝑏, this is equal to root π‘Ž over root 𝑏. So, they’re the same. So therefore, we can say that π‘₯ is equal to plus or minus root seven over root nine. So, π‘₯ is equal to plus or minus root seven over three. And that’s cause root nine is three. So therefore, we can say that the solution to the equation β€” the determinant of the matrix nine π‘₯ squared, two π‘₯, 63π‘₯, nine π‘₯ squared equals negative 49 β€” is going to be the solution set root seven over three and negative root seven over three.

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