
Factorize and solve $25{x^2} - 9 = 0$
Answer
543.9k+ views
Hint:We can use the formula${a^2} - {b^2} = \left( {a + b} \right)\left( {a - b} \right)$. This is the difference of square identity and by using this formula we can factorize the given equation. So in order to factorize we have to convert the question and then express it in the form of the difference of square identity.
Complete step by step solution:
Given
$25{x^2} - 9 = 0................................\left( i \right)$
Also ${a^2} - {b^2} = \left( {a + b} \right)\left( {a - b} \right).....................\left( {ii} \right)$
So we need to express (ii) in terms of (i), for that we have to take the common factors from \[25{x^2} - 9.\]
But we can see that there are no common factors so we can directly express \[25{x^2} - 9\]according to the identity given above.
Such that:
\[ \Rightarrow 25{x^2} - 9 = {\left( {5x} \right)^2} - {\left( 3 \right)^2}......................\left( {iii} \right)\]
Now on comparing (i) and (iii) we get:
\[
\Rightarrow 25{x^2} - 9 = {\left( {5x} \right)^2} - {\left( 3 \right)^2} \\
\Rightarrow {\left( {5x} \right)^2} - {\left( 3 \right)^2} = \left( {5x + 3} \right)\left( {5x - 3}
\right)..............\left( {iv} \right) \\
\]
Therefore on factorization of \[25{x^2} - 9\]we get\[\left( {5x + 3} \right)\left( {5x - 3} \right)\].
Now we have to solve\[25{x^2} - 9\]:
For that we have to consider equation (i), such that:
$
\Rightarrow 25{x^2} - 9 = 0 \\
\Rightarrow \left( {5x + 3} \right)\left( {5x - 3} \right) = 0...........................\left( v \right) \\
$
Now equating each part in (v) to zero, we get:
\[
\Rightarrow 5x + 3 = 0\,\,\,\,\,\,{\text{and}}\,\,\,\,\,\,5x - 3{\text{ = 0}} \\
\Rightarrow x = - \dfrac{3}{5}\,\,\,\,\,\,{\text{and}}\,\,\,\,\,\,x =
\dfrac{3}{5}.............................\left( {vi} \right) \\
\]
Therefore on solving $25{x^2} - 9 = 0$we get \[x = - \dfrac{3}{5}\,{\text{and}}\,x = \dfrac{3}{5}.\]
Additional Information:
Another technique for factoring is grouping, where we take common terms not from the full polynomial but only from the certain terms which have them.
So it is used mainly when there is not a common factor for the whole polynomial but there is a common factor only for certain terms.
However here we use the identity ${a^2} - {b^2} = \left( {a + b} \right)\left( {a - b} \right)$because if we use here grouping then we can’t factorize the expression$25{x^2} - 9.$
Note: While approaching a question one should study it properly and accordingly should choose the method to factorize the polynomial. Similar questions which have real coefficients and some common terms should be approached using the same method as described above. Polynomial factorization is always done over some set of numbers which may be integers, real numbers or complex numbers.
Complete step by step solution:
Given
$25{x^2} - 9 = 0................................\left( i \right)$
Also ${a^2} - {b^2} = \left( {a + b} \right)\left( {a - b} \right).....................\left( {ii} \right)$
So we need to express (ii) in terms of (i), for that we have to take the common factors from \[25{x^2} - 9.\]
But we can see that there are no common factors so we can directly express \[25{x^2} - 9\]according to the identity given above.
Such that:
\[ \Rightarrow 25{x^2} - 9 = {\left( {5x} \right)^2} - {\left( 3 \right)^2}......................\left( {iii} \right)\]
Now on comparing (i) and (iii) we get:
\[
\Rightarrow 25{x^2} - 9 = {\left( {5x} \right)^2} - {\left( 3 \right)^2} \\
\Rightarrow {\left( {5x} \right)^2} - {\left( 3 \right)^2} = \left( {5x + 3} \right)\left( {5x - 3}
\right)..............\left( {iv} \right) \\
\]
Therefore on factorization of \[25{x^2} - 9\]we get\[\left( {5x + 3} \right)\left( {5x - 3} \right)\].
Now we have to solve\[25{x^2} - 9\]:
For that we have to consider equation (i), such that:
$
\Rightarrow 25{x^2} - 9 = 0 \\
\Rightarrow \left( {5x + 3} \right)\left( {5x - 3} \right) = 0...........................\left( v \right) \\
$
Now equating each part in (v) to zero, we get:
\[
\Rightarrow 5x + 3 = 0\,\,\,\,\,\,{\text{and}}\,\,\,\,\,\,5x - 3{\text{ = 0}} \\
\Rightarrow x = - \dfrac{3}{5}\,\,\,\,\,\,{\text{and}}\,\,\,\,\,\,x =
\dfrac{3}{5}.............................\left( {vi} \right) \\
\]
Therefore on solving $25{x^2} - 9 = 0$we get \[x = - \dfrac{3}{5}\,{\text{and}}\,x = \dfrac{3}{5}.\]
Additional Information:
Another technique for factoring is grouping, where we take common terms not from the full polynomial but only from the certain terms which have them.
So it is used mainly when there is not a common factor for the whole polynomial but there is a common factor only for certain terms.
However here we use the identity ${a^2} - {b^2} = \left( {a + b} \right)\left( {a - b} \right)$because if we use here grouping then we can’t factorize the expression$25{x^2} - 9.$
Note: While approaching a question one should study it properly and accordingly should choose the method to factorize the polynomial. Similar questions which have real coefficients and some common terms should be approached using the same method as described above. Polynomial factorization is always done over some set of numbers which may be integers, real numbers or complex numbers.
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