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Repeated roots questions, using the equation from Part a:

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$\\simplify{{f1}x^2+{f2}*x+{f3}}=0$

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These are solved in the exact same way as the previous set of quadratics, but the 'roots' (solutions) are the same.

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This type of equation is described as having 'repeated roots'.

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The graphed functions would only touch the $x$-axis once either at their peak or trough.

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Looking at the above equation, you may be able to instantly see that the brackets must both include the quantity $\\var{c}$.

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If not, don't worry, just go through the same process of trial and error as taught previously until you reach your solution of

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$\\simplify{({a}x+{c})({b}x+{d})}$.

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Then, noticing this can be further simplified, write the final simplification as:

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$\\simplify{({a}x+{c})^2}$.

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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:

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$2x+4=0$  so   $x=\\frac{-4}{2}=-2$

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and

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$x+4=0$    so   $x=-4$

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so are in fact different.

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Zero constants equations, using Part d:

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$\\simplify{{f14}x^2+{f24}*x+{f34}}=0$

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Some equations will not have a constant at the end, meaning one of the constants is zero.

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In this case, if there is still an $x$ term, you know one of the constants must still be non-zero.

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You must therefore, take out a factor of a number $\\times x$.

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Here, the factor could be $\\simplify{{a4}x}$ or $\\simplify{{b4}x}$.

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It doesn't matter which you choose, but in this example we will use $\\simplify{{a4}x}$.

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In order to pull out this factor, the rest of the equation must be divided by $\\simplify{{a4}x}$:

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$\\frac{\\simplify{{f14}x^2+{f24}*x+{f34}}}{\\simplify{{a4}*x}}=\\simplify{{f14}x/{a4}+{f24}/{a4}}$.

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Therefore, the solution is:

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$\\simplify{{a4}x*({f14}x/{a4}+{f24}/{a4})}$.

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$\\simplify{{f1}x^2+{f2}*x+{f3}}$

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The answer must be factorised.

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The answer must be factorised.

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$\\simplify{{f12}x^2+{f22}*x+{f32}}$

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$=$[[0]] 

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The answer must be factorised.

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$\\simplify{{f13}x^2+{f23}*x+{f33}}$

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$=$[[0]] 

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The answer must be factorised.

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$\\simplify{{f14}x^2+{f24}*x+{f34}}$

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$=$[[0]] 

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The answer must be factorised.

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Hint: Remember one or more of the constants could be zero.

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Simplify the following quadratics into two linear factors in the form $(ax+c)(bx+d)$.

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Quadratics 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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Factorising further basic quadratics into linear expressions

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