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By Quadraic formula
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Answer:The quadratic formula helps you solve quadratic equations, and is probably one of the top five formulas in math. We’re not big fans of you memorizing formulas, but this one is useful (and we think you should learn how to derive it as well as use it, but that’s for the second video!).
If you have a general quadratic equation like this:
ax^2+bx+c=0ax
2
+bx+c=0a, x, squared, plus, b, x, plus, c, equals, 0
Then the formula will help you find the roots of a quadratic equation, i.e. the values of xxx where this equation is solved.
The quadratic formula
x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}x=
2a
−b±
b
2
−4ac
x, equals, start fraction, minus, b, plus minus, square root of, b, squared, minus, 4, a, c, end square root, divided by, 2, a, end fraction
It may look a little scary, but you’ll get used to it quickly!
Practice using the formula now.
Worked example
First we need to identify the values for a, b, and c (the coefficients). First step, make sure the equation is in the format from above, ax^2 + bx + c = 0ax
2
+bx+c=0a, x, squared, plus, b, x, plus, c, equals, 0:
x^2+4x-21=0x
2
+4x−21=0x, squared, plus, 4, x, minus, 21, equals, 0
aaa is the coefficient in front of x^2x
2
x, squared, so here a = 1a=1a, equals, 1 (note that aaa can’t equal 000 -- the x^2x
2
x, squared is what makes it a quadratic).
bbb is the coefficient in front of the xxx, so here b = 4b=4b, equals, 4.
ccc is the constant, or the term without any xxx next to it, so here c = -21c=−21c, equals, minus, 21.
Then we plug aaa, bbb, and ccc into the formula:
x=\dfrac{-4\pm\sqrt{16-4\cdot 1\cdot (-21)}}{2}x=
2
−4±
16−4⋅1⋅(−21)
x, equals, start fraction, minus, 4, plus minus, square root of, 16, minus, 4, dot, 1, dot, left parenthesis, minus, 21, right parenthesis, end square root, divided by, 2, end fraction
solving this looks like:
\begin{aligned} x&=\dfrac{-4\pm\sqrt{100}}{2} \\\\ &=\dfrac{-4\pm 10}{2} \\\\ &=-2\pm 5 \end{aligned}
x
=
2
−4±
100
=
2
−4±10
=−2±5
Therefore x = 3x=3x, equals, 3 or x = -7x=−7x, equals, minus, 7.
What does the solution tell us?
The two solutions are the x-intercepts of the equation, i.e. where the curve crosses the x-axis. The equation x^2 + 3x - 4 = 0x
2
+3x−4=0x, squared, plus, 3, x, minus, 4, equals, 0 looks like:
Graphing quadratic equations
Graphing quadratic equations
where the solutions to the quadratic formula, and the intercepts are x = -4x=−4x, equals, minus, 4 and x = 1x=1x, equals, 1.
Now you can also solve a quadratic equation through factoring, completing the square, or graphing, so why do we need the formula?
Because sometimes quadratic equations are a lot harder to solve than that first example.
Second worked example
Let’s try this for an equation that is hard to factor:
3x^2+6x=-103x
2
+6x=−103, x, squared, plus, 6, x, equals, minus, 10
Let’s first get it into the form where all terms are on the left-hand side:
\underbrace{(3)}_{a}x^2+\underbrace{(6)}_{b}x+\underbrace{(10)}_{c}=0
a
(3)
x
2
+
b
(6)
x+
c
(10)
=0start underbrace, left parenthesis, 3, right parenthesis, end underbrace, start subscript, a, end subscript, x, squared, plus, start underbrace, left parenthesis, 6, right parenthesis, end underbrace, start subscript, b, end subscript, x, plus, start underbrace, left parenthesis, 10, right parenthesis, end underbrace, start subscript, c, end subscript, equals, 0
The formula gives us:
\begin{aligned} x&=\dfrac{-6\pm\sqrt{6^2-4\cdot 3\cdot 10}}{2\cdot 3} \\\\ &=\dfrac{-6\pm\sqrt{36-120}}{6} \\\\ &=\dfrac{-6\pm\sqrt{-84}}{6} \end{aligned}
x
=
2⋅3
−6±
6
2
−4⋅3⋅10
=
6
−6±
36−120
=
6
−6±
−84
We know you can’t take the square root of a negative number without using imaginary numbers, so that tells us there’s no real solutions to this equation. This means that at no point will y = 0y=0y, equals, 0, the function won’t intercept the x-axis.
Be careful that the equation is arranged in the right form: ax^2 + bx + c = 0ax
2
+bx+c=0a, x, squared, plus, b, x, plus, c, equals, 0 or it won’t work!
Make sure you take the square root of the whole (b^2 - 4ac)(b
2
−4ac)left parenthesis, b, squared, minus, 4, a, c, right parenthesis, and that 2a2a2, a is the denominator of everything above it
Watch your negatives: b^2b
2
b, squared can’t be negative, so if bbb starts as negative, make sure it changes to a positive since the square of a negative or a positive is a positive
Keep the +/-+/−plus, slash, minus and always be on the look out for TWO solutions
If you use a calculator, the answer might be rounded to a certain number of decimal places. If asked for the exact answer (as usually happens) and the square roots can’t be easily simplified, keep the square roots in the answer
Step-by-step explanation: