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Repeated roots questions, using the equation from Part a:
\n$\\simplify{{f1}x^2+{f2}*x+{f3}}=0$
\nThese are solved in the exact same way as the previous set of quadratics, but the 'roots' (solutions) are the same.
\nThis type of equation is described as having 'repeated roots'.
\nThe graphed functions would only touch the $x$-axis once either at their peak or trough.
\nLooking at the above equation, you may be able to instantly see that the brackets must both include the quantity $\\var{c}$.
\nIf not, don't worry, just go through the same process of trial and error as taught previously until you reach your solution of
\n$\\simplify{({a}x+{c})({b}x+{d})}$.
\nThen, noticing this can be further simplified, write the final simplification as:
\n$\\simplify{({a}x+{c})^2}$.
\n\n\n
\n
Note: If the $x^2$ term has a coefficient and simplifies to, for example, $(2x+4)(x+4)$, this is not defined as having repeated roots, since the solutions are:
\n$2x+4=0$ so $x=\\frac{-4}{2}=-2$
\nand
\n$x+4=0$ so $x=-4$
\nso are in fact different.
\n\n\n\nZero constants equations, using Part d:
\n$\\simplify{{f14}x^2+{f24}*x+{f34}}=0$
\nSome equations will not have a constant at the end, meaning one of the constants is zero.
\nIn this case, if there is still an $x$ term, you know one of the constants must still be non-zero.
\nYou must therefore, take out a factor of a number $\\times x$.
\nHere, the factor could be $\\simplify{{a4}x}$ or $\\simplify{{b4}x}$.
\nIt doesn't matter which you choose, but in this example we will use $\\simplify{{a4}x}$.
\nIn order to pull out this factor, the rest of the equation must be divided by $\\simplify{{a4}x}$:
\n$\\frac{\\simplify{{f14}x^2+{f24}*x+{f34}}}{\\simplify{{a4}*x}}=\\simplify{{f14}x/{a4}+{f24}/{a4}}$.
\nTherefore, the solution is:
\n$\\simplify{{a4}x*({f14}x/{a4}+{f24}/{a4})}$.
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\nQuadratics are defined as having 'repeated roots' if the both roots in the solution are the same. This can be written as $(x+a)(x+a)$, but can be simplified to $(x+a)^2$.
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