The quadratic equation below has x2-4 root 3x-3=0
Answers
Answer:
do some more examples.
Solve x (x – 2) = 4. Round your answer to two decimal places.
I not only cannot apply the Quadratic Formula at this point, I cannot factor either. Why? Because this equation is yet in the correct form.
And I certainly can not claim, with a straight face, that "x = 4, x – 2 = 4", because this is not how "solving by factoring" works.
No matter which solution method I intend to use — whether I'm factoring or using the Quadratic Formula to find my answers — I must first rearrange the equation into the form "(quadratic) = 0".
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The Quadratic Formula
The first thing I'll do here is multiply through on the left-hand side, and then I'll move the 4 over from the right-hand side to the left-hand side:
x (x – 2) = 4
x2 – 2x = 4
x2 – 2x – 4 = 0
Since there are no factors of (1)(–4) = –4 that add up to –2, then this quadratic does not factor. (In other words, there is no possible way that the faux-factoring solution of "x = 4, x – 2 = 4" could be even slightly correct.)
So factoring won't work, but I can use the Quadratic Formula; in this case, I'll be plugging in the values a = 1, b = –2, and c = –4:
x = \dfrac{-(-2) \pm \sqrt{(-2)^2 - 4(1)(-4)\,}}{2(1)}x=
2(1)
−(−2)±
(−2)
2
−4(1)(−4)
= \dfrac{2 \pm \sqrt{4 + 16\,}}{2} = \dfrac{2 \pm \sqrt{20\,}}{2}=
2
2±
4+16
=
2
2±
20
= \dfrac{2 \pm \sqrt{4\,}\sqrt{5\,}}{2} = \dfrac{2 \pm 2\, \sqrt{5\,}}{2}=
2
2±
4
5
=
2
2±2
5
= \dfrac{2\,\left(1 \pm \sqrt{5\,}\right)}{2(1)} = 1 \pm \sqrt{5\,}=
2(1)
2(1±
5
)
=1±
5
\approx 1.236068,\, 3.236968≈1.236068,3.236968
Then the answer is:
x = –1.24, x = 3.24, rounded to two decimal places.
For reference, here's what the graph of the associated quadratic, y = x2 – 2x – 4, looks like:
y = x^2 - 2x - 4
As you can see, the solutions from the Quadratic Formula match up with the x-intercepts. The locations where the graph crosses the x-axis give the values that solve the original equation.
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There is another connection between the solutions from the Quadratic Formula and the graph of the parabola: you can tell how many x-intercepts you're going to have from the value inside the square root. The argument (that is, the contents) of the square root, being the expression b2 – 4ac, is called the "discriminant" because, by using its value, you can "discriminate" between (that is, be able to tell the difference between) the various solution types.
In this case, the value of the discriminant b2 – 4ac was 20; in particular, the value was not zero and was not negative. Because the value was not negative, the equation was going to have at least one (real-valued) solution; because the value was not zero, the two solutions were going to be distinct (that is, they were going to be different from each other).
Solve 9x2 + 12x + 4 = 0. Leave your answer in exact form.
Using a = 9, b = 12, and c = 4, the Quadratic Formula gives me:
x = \dfrac{-(12) \pm \sqrt{(12)^2 - 4(9)(4)\,}}{2(9)}x=
2(9)
−(12)±
(12)
2
−4(9)(4)
= \dfrac{-12 \pm \sqrt{144 - 144\,}}{18}=
18
−12±
144−144
= \dfrac{-12 \pm \sqrt{0\,}}{18} = \dfrac{-12 \pm 0}{18}=
18
−12±
0
=
18
−12±0
= \dfrac{-12}{18} = \dfrac{-2}{3} = -\dfrac{2}{3}=
18
−12
=
3
−2
=−
3
2
Then the answer is:
\mathbf{\color{purple}{\mathit{x} = -\dfrac{2}{3}}}x=−
3
2
In the first example on this page, I had gotten two solutions because the value of the discriminant (that is, the value inside the square root) was non-zero and positive. As a result, the "plus-minus" part of the Formula gave me two distinct values; one for the "plus" part of the numerator and another for the "minus" part. In this case, though, the square root reduced to zero, so the plus-minus didn't count for anything.
This sort of solution, where you get only one value because "plus or minus zero" didn't change anything, is called a "repeated" root, because x is equal to - \frac{2}{3}−
3
2
, but it's equal to this value kind-of twice: - \frac{2}
hope it will help you{3} + 0−
3
2
+0 and - \frac{2}{3} - 0−
3
2
−0.