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Apply the quadratic formula to find the roots of a given equation. The quadratic formula is given in the steps if the student requires it.

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An equation of the form

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\$ax^2+bx+c=0\\text{,}\$

\n

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can be solved using the quadratic formula

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\$x={\\frac {-b\\pm\\sqrt{b^2-4\\times a\\times c}}{2a}}\\text{.}\$

\n

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$\\simplify{x^2+{a+m}x+{a*m}=0}$

\n

$x_1=$ [[0]]

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$x_2=$ [[1]]

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\n

$\\simplify{{a1}x^2+{a2}x+{a3}={a4}}$

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$x_1=$ [[0]]

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$x_2=$ [[1]]

\n

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$\\simplify{{b1}x^2+{b2}x+{b3}={b4}x}$

\n

$x_1=$ [[0]]

\n

$x_2=$ [[1]]

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When quadratic equations can't be factorised, or if equations are difficult to factorise (perhaps if the coefficients are large), we need to use the quadratic formula to solve the equations.

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Use the quadratic formula to calculate values for $x$ in these equations. Input the possible values as $x_1$ and $x_2$, where $x_1<x_2$.

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\$x={\\frac {-b\\pm\\sqrt{b^2-4\\times a\\times c}}{2a}}\\text{.}\$

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#### a)

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From the equation, we can read off values for $a$, $b$ and $c$:

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\\\begin{align} a&=1\\text{,}\\\\ b&=\\var{a+m}\\text{,}\\\\ c&=\\var{a*m} \\text{.} \\end{align}\

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Substituting these values into the quadratic formula,

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\$x = \\frac {-\\var{a+m}\\pm\\sqrt{\\var{a+m}^2-4\\times \\var{a*m}}}{2}\\text{.}\$

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Note the $\\pm$ symbol in the formula. This means there are two solutions: one using $+$, the other using $-$.

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The two solutions are

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\\\begin{align} x_1&=\\var{m}\\text{,}\\\\ x_2&=\\var{a}\\text{.} \\end{align}\

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#### b)

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Note that the right-hand side of the given equation is not zero. We need to rewrite it in the form $ax^2+bx+c=0$:

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\\\begin{align} \\simplify{{a1}x^2+{a2}x+{a3}}&=\\var{a4}\\\\ \\simplify{{a1}x^2+{a2}x+{a3-a4}}&=0\\text{.} \\end{align}\

\n

Then we can read off values for $a$, $b$ and $c$:

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\\\begin{align} a&=\\var{a1}\\\\ b&=\\var{a2}\\\\ c&=\\var{a3-a4} \\text{.} \\end{align}\

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We can now substitute these values into the quadratic formula:

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\$x = {\\frac {-\\var{a2}\\pm\\sqrt{\\var{a2}^2-4\\times \\var{a1}\\times \\var{a3-a4}}}{2\\times\\var{a1}}}\\text{.}\$

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So the two solutions are

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\\\begin{align} x_1&=\\var{dpformat(x1,2)}\\\\ x_2&=\\var{dpformat(x2,2)}\\text{.} \\end{align}\

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#### c)

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We first rearrange our equation into the form $ax^2+bx+c=0$:

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\\\begin{align} \\simplify{{b1}x^2+{b2}x+{b3}}&=0=\\var{b4}x\\\\ \\simplify{{b1}x^2+{b2-b4}x+{b3}}&=0\\text{.} \\end{align}\

\n

We can then read off the values for $a, b$ and $c$, which are

\n

\\\begin{align} a&=\\var{b1}\\text{,}\\\\ b&=\\var{b2-b4}\\text{,}\\\\ c&=\\var{b3}\\text{.} \\end{align}\

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Substituting these values into the quadratic formula,

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\$x = {\\frac {-\\var{b2-b4}\\pm\\sqrt{\\var{b2-b4}^2-4\\times \\var{b1}\\times \\var{b3}}}{2\\times\\var{b1}}},\$

\n

we obtain solutions

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\\\begin{align} x_1&=\\var{dpformat(p1,2)}\\text{,}\\\\ x_2&=\\var{dpformat(p2,2)}\\text{.} \\end{align}\

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