Video Transcript
A student correctly substitutes
values into a quadratic formula as follows: 𝑥 equals five plus or minus the square
root of 25 minus 80 all over eight. What is the quadratic equation they
were trying to solve?
In this problem, we’ve been told
that a student has used a quadratic formula to attempt to solve a quadratic
equation. We’re given their solution partway
through. And we’re also told that their
working out is correct. Now the reason the question asked
for the equation the student was trying to solve is because if we look closely, we
can see that the value underneath the square root will be negative. It’s 25 minus 80 which is equal to
negative 55. The student is trying to find the
square root of a negative value, which we know we can’t do in the real numbers. This doesn’t mean their working out
is wrong. It just means that there are no
real solutions to this equation.
In order to work out the quadratic
equation they were trying to solve, we’re going to need to work backwards. Let’s recall the quadratic
formula. This tells us that the roots of the
general quadratic equation 𝑎𝑥 squared plus 𝑏𝑥 plus 𝑐 equals zero, where 𝑎, 𝑏,
and 𝑐 are real constants and 𝑎 is nonzero, are given by 𝑥 equals negative 𝑏 plus
or minus the square root of 𝑏 squared minus four 𝑎𝑐 all over two 𝑎. We therefore need to work out what
the values of 𝑎, 𝑏, and 𝑐 are for this quadratic equation, as this will give us
the coefficients of 𝑥 squared, 𝑥, and the constant term. We can do this by comparing the
different parts of the student’s solution with the corresponding parts in the
general quadratic formula.
Let’s compare the denominators of
the two fractions first of all. In the general quadratic formula,
we have two 𝑎, whereas in the student’s solution, we have eight. So this gives the equation two 𝑎
is equal to eight. We can solve this equation by
dividing both sides by two, and it gives 𝑎 is equal to four. We’ve therefore found the values of
one of the three unknowns 𝑎, 𝑏, and 𝑐.
Next, let’s look at the first term
in the numerator. In the general quadratic formula,
this is given by negative 𝑏. And in the student’s solution, this
is given by five. Equating these, we have the
equation negative 𝑏 is equal to five. We can solve this equation by
either multiplying or dividing both sides by negative one. And it gives 𝑏 is equal to
negative five. So we’ve now found two of the
unknowns. And in fact, if we look under the
square root, we can see that the first term under the square root is 𝑏 squared. And this is equal to 25 in the
student’s solution. If we take our value of negative
five for 𝑏 and square it, we do indeed get 25. So this confirms that our value of
𝑏 is correct.
Let’s now compare the other part
under the square root. We have 𝑏 squared, and then the
part that follows it is minus four 𝑎𝑐 or negative four 𝑎𝑐. In the student’s solution, this is
equivalent to negative 80. So we have the equation negative
four 𝑎𝑐 equals negative 80. And in fact, we can multiply both
sides of this equation by negative one to simplify. So we have simply four 𝑎𝑐 equals
80. Remember, we’ve already worked out
the value of 𝑎. 𝑎 is equal to four. So substituting 𝑎 equals four, on
the left-hand side, we have four multiplied by four multiplied by 𝑐; that’s
16𝑐. So our equation becomes 16𝑐 is
equal to 80.
To solve, we divide both sides of
the equation by 16, giving 𝑐 equals 80 over 16, which simplifies to five. We’ve therefore found the values of
the constants 𝑎, 𝑏, and 𝑐. All that’s left to do is to
substitute these into the general form of a quadratic equation. Remember, 𝑎 represents the
coefficient of 𝑥 squared, 𝑏 represents the coefficient of 𝑥, and 𝑐 represents
the constant term. So we have the equation four 𝑥
squared minus five 𝑥 plus five is equal to zero. And this is the quadratic equation
the student was trying to solve by applying the quadratic formula.
