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The quadratic formula is

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\\[{\\frac {-b\\pm\\sqrt{b^2-4\\times a\\times c}}{2a}}\\text{.}\\]

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Solve the equation $\\simplify {x^2+{b_3}*k*x+{c_2}k^2=0}$. Give your answer in terms of $k$. Assuming $k$ is positive, enter the lowest root first.

\n

$x_1=$ [[0]]

\n

$x_2=$ [[1]]

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Factorise a quadratic expression of the form $x^2+akx+bk^2$ for $x$, in terms of $k$. $a$ and $b$ are constants.

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The coefficients of the following equation involve the unknown value $k$. We can use the quadratic formula to find expressions for the values of $x$ in terms of $k$.

", "advice": "

The quadratic formula is

\n

\\[{\\frac {-b\\pm\\sqrt{b^2-4\\times a\\times c}}{2a}}\\text{.}\\]

\n

We can list our values for $a, b$ and $c$.

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\\[\\begin{align}
a&=1\\\\
b&=\\var{b_3}k\\\\
c&=\\var{c_2}k^2
\\end{align}\\]

\n

Then by substituting them into the quadratic formula, we obtain

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\\[x=\\frac {-\\var{b_3}k\\pm\\sqrt{\\var{b_3}^2k^2-4\\times \\var{c_2}k^2}}{2}\\]

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We can then simplify this equation to

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\\[\\begin{align}
x&=\\frac {-\\var{b_3}k\\pm k\\sqrt{\\var{b_3}^2-\\var{4c_2}}}{2}\\\\
\\end{align}\\]
\\[\\begin{align}
&=k\\left(\\frac{-\\var{b_3}}{2}\\pm \\frac {\\sqrt{\\var{b_3}^2-\\var{4c_2}}}{2}\\right)\\\\
\\end{align}\\]
\\[\\begin{align}
&=k\\left(-\\frac{\\var{b_3}}{2} \\pm\\frac{\\sqrt{\\var{(b_3^2-4c_2)}}}{2}\\right)\\\\
\\end{align}\\]

\n

This means our possible values for $x$ in terms of $k$ are,

\n

\\begin{align}
x_1 &= \\left( \\simplify[all,!noleadingminus,!collectnumbers,!simplifyfractions]{-{b_3}/2 - {sqrt(b_3^2-4c_2)}/2} \\right) k = \\var[fractionnumbers]{-b_3/2 - sqrt(b_3^2-4*c_2)/2}k \\\\
x_2 &= \\left( \\simplify[all,!noleadingminus,!collectnumbers,!simplifyfractions]{-{b_3}/2 + {sqrt(b_3^2-4c_2)}/2} \\right) k = \\var[fractionnumbers]{-b_3/2 + sqrt(b_3^2-4*c_2)/2}k
\\end{align}

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