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# What is the standard form of $y = {\left( {2x - 3} \right)^2}$?

Last updated date: 22nd Jul 2024
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Hint: We can see that this is the quadratic equation. But this is not in its standard form which is given as: $y = a{x^2} + bx + c$. So, to convert it into its standard form we will expand the term in the RHS using the formula for ${\left( {a - b} \right)^2}$.
Formula used:
${\left( {a - b} \right)^2} = {a^2} - 2ab + {b^2}$

Given equation:
$y = {\left( {2x - 3} \right)^2}$
Now to get it in the standard form we will expand the RHS term using the formula ${\left( {a - b} \right)^2} = {a^2} - 2ab + {b^2}$. So, we get;
$\Rightarrow y = {\left( {2x} \right)^2} - 2\left( {2x} \right)3 + {3^2}$
Solving we get;
$\Rightarrow y = 4{x^2} - 12x + 9$
This is in the standard form i.e., $y = a{x^2} + bx + c$, with $a = 4,b = - 12,c = 9$.

Note:
For the above equation we can find the roots by equating the equation to zero. So, we have;
$y = 4{x^2} - 12x + 9 = 0$
$\Rightarrow 4{x^2} - 12x + 9 = 0$
Splitting the middle term, we get;
$\Rightarrow 4{x^2} - 6x - 6x + 9 = 0$
Taking the common we get;
$\Rightarrow 2x\left( {2x - 3} \right) - 3\left( {2x - 3} \right) = 0$
Further taking common we get;
$\Rightarrow \left( {2x - 3} \right)\left( {2x - 3} \right) = 0$
$\Rightarrow x = \dfrac{2}{3}$