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BloomLocke367

  • one year ago

Tutorial: How to solve quadratic equations for x.

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  1. BloomLocke367
    • one year ago
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    To be able to solve quadratic equations, you first have to know what a quadratic equation is. A quadratic equation is in the form of \(\large\color{red}ax^2\small+\large\color{blue}bx\small+\large\color{gold}c\), where \(\large\color{red}a\neq0\). Quadratic equations generally have two solutions, but one may repeat, or there might not be a solution at all.

  2. BloomLocke367
    • one year ago
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    There are four ways to solve for x. The first is to solve by factoring. To do this, you have to find what two numbers multiply together to get ac, but add up to get b. Here are some examples:

  3. BloomLocke367
    • one year ago
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    \(\large x^2+x-\color{gold}6=0\) Notice that a and b are both one. Step 1: Find the two numbers that multiply to get -6, but add to get 1. \(\large3+(-2)=1\) Step 2: Write it in factored form. \(\large(x+3)(x-2)=0\) Step 3: Break it up into two parts in order to solve for x. \(\large x+3=0~and~x-2=0\) Step 4: Solve for x! \(\large x=-3 ~and~ x=2\) you can condense your answer to make it easier to read by stating \(\large x=-3, ~2\) \(\large \color{red}2x^2\small-\large\color{blue}7x\small+\large\color{gold}3=0\) This one is a little different because the coefficient is not 1. Step 1: Multiply a and c. \(\large\color{red}2\small\times\large\color{gold}3=6\) Step 2: Find the two factors of 6 that add up to -7. \(\small-\large6\small+\large(\small-\large1)=\small-\large7.\) Step 3: Split the middle term using the newfound factors \(\large\color{red}{ 2x^2\small-\large6x}\color{green}{\small-\large1x\small+\large3=0}\) Step 4: Pull out the factors of the first half (the first two terms). \(\color{red}{\large 2x\small\times\large(x\small-\large3)}\) Step 4: Pull out the factors of the second half (the last two terms). \(\color{green}{\large1\small\times\large(x\small-\large3)}\) Step 5: Add the terms found in steps 3 and 4. \(\color{hotpink}{\large(2x\small+\large1)\small\times\large(x\small-\large3)}\) Step 6: Break it up. \(\color{hotpink}{\large2x\small+\large1=0}\) and \(\color{hotpink}{\large x\small-\large3=0}\) Step 7: Solve for x! \(\color{hotpink}{\large2x=\small-\large1}\) and \(\color{hotpink}{\large x\small=3}\) \(\color{hotpink}{\large x=\small-\large\frac{1}{2}}\) Step 8: Rewrite your final answer \(\color{hotpink}{\large x=\small-\large\frac{1}{2}, ~ 3}\)

  4. BloomLocke367
    • one year ago
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    The second way to solve for x is to graph it. All you need to know is how to read the graph that comes up. Quadratic functions create a parabola, which is a U shape. When you graph a quadratic, \(\large\color{gold}c\) would be the y-intercept. That is where the point of the parabola will cross the y-axis. Your solutions, if there are any, would be the x-intercepts, or where the graph passes through the x-axis.

  5. BloomLocke367
    • one year ago
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    The third way to solve for x is to use the quadratic formula. This is the guaranteed way to find the solutions, when the other options are only good for certain equations. The quadratic formula is \(\Large\frac{x=\small-\Large\color{blue} b\pm\sqrt{\color{blue}b^2\small-\Large4\color{red}a\color{gold}c}}{2\color{red}a}\)

  6. BloomLocke367
    • one year ago
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    Here is an example of how to use the quadratic formula: \(\large x^2\small-\large\color{blue}2x\small-\large\color{gold}8\). (there is an understood 1 as the coeffecient, or \(\color{red} a\).) Step 1: Plug in the values into the formula. \(\LARGE x=\frac{−(\color{blue}{-2})\pm\sqrt{(\color{blue}{−2})^2−4(\color{red}1)(\color{gold}{−8})}}{2(\color{red}1)}\) Step 2: Simplify the numerator (what's on top). \(\LARGE x=\frac{\color{blue}{2}\pm\sqrt{\color{green}4−4(\color{red}1)(\color{gold}{−8})}}{2(\color{red}1)}\) \(\LARGE x=\frac{\color{blue}{2}\pm\sqrt{\color{green}{4+32}}}{2(\color{red}1)}\) \(\LARGE x=\frac{\color{blue}{2}\pm\sqrt{\color{green}{36}}}{2(\color{red}1)}\) \(\LARGE x=\frac{\color{blue}{2}\pm\color{green} 6}{2(\color{red}1)}\) Step 3: Simplify the denominator (the bottom part of the fraction). \(\LARGE x=\frac{\color{blue}{2}\pm\color{green} 6}{2}\) Step 4: Simplify the entire fraction (take the common factor out of the top and bottom). \(\LARGE x=\frac{\color{blue}{1}\pm\color{green} 3}{1}\) \(\large x=\color{blue}{1}\pm\color{green} 3\) Step 5: Split it up into two equations. \(\large x=\color{hotpink}{1+3}\) and \(\large x=\color{hotpink}{1−3} \) Step 6: Solve! \(\large x=4\) and \(\large x=-2\) Step 7: Condense your answer. \(\large x=-2,4\) You may not always get a real number as an answer, but I'll explain that in my next tutorial on imaginary and complex numbers.

  7. BloomLocke367
    • one year ago
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    The final way to solve for quadratics is to use the method called Completing the Square. This method may be a little tricky at first, but with practice, it becomes quite easy! Here's an example:

  8. BloomLocke367
    • one year ago
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    \(\large\color{red}4x^2-\color{blue}2x-\color{gold}5=0\) Step 1: Get the constant, or \(\color{gold}c\) to the other side of the equation. \(\large\color{red}4x^2-\color{blue}2x=\color{gold}5\) Step 2: Divide everything by \(\color{red}a\). \(\large x^2-\LARGE\frac{\color{blue}2}{\color{red}{4}}x=\LARGE\frac{\color{gold}5}{\color{red}{4}}\) \(\large x^2-\LARGE\color{green}{\frac{1}{2}}x=\LARGE\color{green}{\frac{5}{4}}\) Step 3: Take half of the new \(\color{blue}b\) (don't forget to include the sign). \(\LARGE\color{green}{-\frac{1}{2}}\large\div 2=\LARGE\color{hotpink}{-\frac{1}{4}}\) Step 4: Square the new number. \(\LARGE\color{hotpink}{-\frac{1}{4}}^2=\frac{1}{16}\) Step 5: Add the squared number to both sides of the equation. \(\large\color{red}{4}\color{black}{x^2-\LARGE \color{green}{\frac{1}{2}}+\color{teal}{\frac{1}{16}}=\color{green}{\frac{5}{4}}+\color{teal}{\frac{1}{16}}}\) Step 6: Factor the left hand side. \(\large (x-\LARGE\color{hotpink}{\frac{1}{4}}\large)^2=\LARGE\color{green}{\frac{5}{4}}+\color{teal}{\frac{1}{16}}\) Step 7: Simplify the right hand side. \(\large (x-\LARGE\color{hotpink}{\frac{1}{4}}\large)^2=\LARGE\color{purple}{\frac{21}{16}}\) Step 8: Take the square root of both sides (don't forget it's plus or minus on the right side). \(\large x-\LARGE\color{hotpink}{\frac{1}{4}}\large=\pm\sqrt{\LARGE\color{purple}{\frac{21}{16}}}\) Step 9: Break the radical up. This is allowed because of the quotient rule of radicals, and must be done to further solve the equation. \(\large x-\LARGE\color{hotpink}{\frac{1}{4}}\large=\pm \LARGE\color{purple}{\frac{\sqrt{21}}{\sqrt{16}}}\) Step 10: Simply the radical. \(\large x-\LARGE\color{hotpink}{\frac{1}{4}}\large=\pm \LARGE\color{purple}{\frac{\sqrt{21}}{4}}\) Step 11: Solve for x. \(\large x=\LARGE\color{hotpink}{\frac{1}{4}}\large\pm \LARGE\color{purple}{\frac{\sqrt{21}}{4}}\) \(\large x=\LARGE\color{hotpink}{\frac{1}{4}}\large+\LARGE\color{purple}{\frac{\sqrt{21}}{4}}\) and \(\large x=\LARGE\color{hotpink}{\frac{1}{4}}\large-\LARGE\color{purple}{\frac{\sqrt{21}}{4}}\) Step 12: Condense your answer. \(\large x=\LARGE \frac{1}{4}\large+ \LARGE{\frac{\sqrt{21}}{4}}\), \(\large x=\LARGE{\frac{1}{4}}\large- \LARGE{\frac{\sqrt{21}}{4}}\)

  9. BloomLocke367
    • one year ago
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    That's it for solving quadratic equations! The quadratic equation will always be useful to get your answers, but it may not always be the easiest. It just depends on the type of equation, but with practice, you'll be a master at all of the different ways to solve them!

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