Video: Applying Knowledge of the Quadratic Formula

Which expression indicates both solutions of the equation 2π‘₯Β² + 10π‘₯ βˆ’ 7 = 0? [A] (βˆ’5 Β± √(39))/2 [B] (βˆ’5 Β± π‘–βˆš(44))/2 [C] (βˆ’5 Β± 2√(39))/2 [D] (βˆ’5 Β± π‘–βˆš(11))/2

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Video Transcript

Which expression indicates both solutions of the equation two π‘₯ squared plus 10π‘₯ minus seven equals zero? Is it A) negative five plus or minus the square root of 39 all over two? B) Negative five plus or minus 𝑖 times the square root of 44 all over two? C) Negative five plus or minus two times the square root of 39 all over two? Or D) negative five plus or minus 𝑖 times the square root of 11 all over two?

So, the question tells us that there are two values of π‘₯ that satisfy this equation two π‘₯ squared plus 10π‘₯ minus seven equals zero. And we’ve got to come up with a single expression that encapsulates both of those values of π‘₯. Now, you should recognise that this equation is a quadratic equation. We’ve got a quadratic expression two π‘₯ squared plus 10π‘₯ minus seven equal to zero.

And the quadratic formula tells us that when we’ve got the general form of a quadratic equation like π‘Žπ‘₯ squared plus 𝑏π‘₯ plus 𝑐 is equal to zero, then π‘₯ is equal to the negative of 𝑏 plus or minus the square root of 𝑏 squared minus four π‘Žπ‘ all over two times π‘Ž. So, we need to identify the values for π‘Ž, 𝑏, and 𝑐 and plug them into this formula. And in our equation, the coefficient of π‘₯ squared is two, so π‘Ž is equal to two. The coefficient of π‘₯ is positive 10, so 𝑏 is equal to 10. And the constant term is negative seven, so 𝑐 is equal to negative seven.

Now, just before we go through and substitute in the values for π‘Ž, 𝑏, and 𝑐 into our formula, let’s just remember that four π‘Žπ‘ means four times π‘Ž times 𝑐, and two π‘Ž means two times π‘Ž. So, 𝑏 is equal to 10. So, negative 𝑏 is equal to negative 10. And 𝑏 squared is 10 squared. π‘Ž is two. And 𝑐 is negative seven. So, negative four π‘Žπ‘ becomes negative four times two times negative seven. And a top tip is to write negative numbers in parentheses when you’re substituting them into these formulae. Otherwise you might end up subtracting seven rather than multiplying by negative seven.

And on the denominator, remember, π‘Ž was equal to two. So, two π‘Ž becomes two times two. So, now, we know π‘₯ is equal to negative 10 plus or minus the square root of 10 squared minus four times two times negative seven all over two times two. Now, taking a quick look at the options we were given for answers, looks like we’re gonna have to do a bit of simplifying. Well, on the denominator, two times two is equal to four. 10 squared is equal to 100. Negative four times two is negative eight. And negative eight times negative seven is positive 56. And now, 100 plus 56 is 156. So, now, we’ve simplified, but it still doesn’t look like any of the options that we’ve got.

All the options we’re looking for have a denominator of two, and the numerator starts with negative five. But that denominator is four. And our numerator starts with negative 10. If we could take out two as a factor and cancel this down, we might be able to get closer to the answer we’re looking for. Now, that’s gonna mean doing something with this number under the square root. Now, we can express negative 10 as two times negative five. And we can express four as two times two.

So, if we can express the square root of 156 as two times the square root of something else, then we’ll be able to take a factor of two on the numerator and we’ll be able to cancel that with a factor of two on the denominator. Now, two squared is four. So, two times the square root of something is the same as the square root of four times the square root of that thing. And we can express that as the square root of four times the thing. So, if we divide 156 by four, we get 39. So, the square root of 156 is the same as the square root of four times 39.

And that is the same as the square root of four times the square root of 39. And that is the same as two times the square root of 39. So, bringing that altogether, we’ve got π‘₯ is equal to two times negative five plus or minus two times the square root of 39 all over two times two. The numerator has a common factor of two, so we can factor that out. Now, we’ve got two times negative five plus or minus the square root of 39 on the numerator and two times two on the denominator. If we divide both by two, these twos cancel. And we end up with just negative five plus or minus the square root of 39 all over two.

So, our answer is A, negative five plus or minus the square root of 39 all over two. Now, that just means if we take π‘₯ to be equal to negative five plus square root 39 all over two or we take π‘₯ to be equal to negative five minus root 39 all over two either of those values of π‘₯ will satisfy our original equation. Take either of those values for π‘₯ and substitute them in here and here to original equation, and you will indeed get an answer of zero.

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