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

seo-qna
Last updated date: 22nd Jul 2024
Total views: 324k
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Answer
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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:
\[{\left( {a - b} \right)^2} = {a^2} - 2ab + {b^2}\]

Complete step by step answer:
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;
\[ \Rightarrow \left( {2x - 3} \right)\left( {2x - 3} \right) = 0\]
\[ \Rightarrow x = \dfrac{2}{3}\]