
How do you solve \[{{x}^{2}}-48=0\]?
Answer
564k+ views
Hint: Write 48 as the product of its prime factors and if any factor is repeated then write it in the exponent form. Try to convert the factors as their exponent equal to 2. Now, apply the algebraic identity: - \[{{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)\] to simplify the given quadratic expression and substitute each term equal to 0 to get the two roots of the equation.
Complete step by step answer:
Here, we have been provided with the quadratic expression: - \[{{x}^{2}}-48=0\] and we are asked to solve it. That means we have to find the values of x.
Now, as we can see that the given equation is quadratic in nature, so we must have two roots or values of x. Since, the coefficient of x in the above equation is 0, so we will not apply the middle term split method. Here, let us apply the factorization method using the algebraic identity: - \[{{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)\].
Now, writing 48 as the product of its primes, we get,
\[\Rightarrow 48=2\times 2\times 2\times 2\times 3\]
This can be written as: -
\[\begin{align}
& \Rightarrow 48={{2}^{2}}\times {{2}^{2}}\times 3 \\
& \Rightarrow 48={{\left( 2\times 2 \right)}^{2}}\times 3 \\
& \Rightarrow 48={{4}^{2}}\times 3 \\
\end{align}\]
Here, we need to convert the factor 3 such that its exponent becomes 2 and we can apply the algebraic identity. To do this we will take the square root of 3 and to balance we will square it. So, we have,
\[\begin{align}
& \Rightarrow 48={{4}^{2}}\times {{\left( \sqrt{3} \right)}^{2}} \\
& \Rightarrow 48={{\left( 4\sqrt{3} \right)}^{2}} \\
\end{align}\]
Substituting this converted form of 48 in the quadratic equation: -
\[{{x}^{2}}-48\], we get,
\[\Rightarrow {{x}^{2}}-{{\left( 4\sqrt{3} \right)}^{2}}=0\]
Using the algebraic identity: - \[{{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)\] we get,
\[\Rightarrow \left( x+4\sqrt{3} \right)\left( x-4\sqrt{3} \right)=0\]
Substituting each term equal to 0, we get,
\[\Rightarrow \left( x+4\sqrt{3} \right)=0\] or \[\left( x-4\sqrt{3} \right)=0\]
\[\Rightarrow x=-4\sqrt{3}\] or \[x=4\sqrt{3}\]
Hence, the roots of the given quadratic equation are: - \[4\sqrt{3}\] and \[-4\sqrt{3}\].
Note:
One may note that whatever method we may apply we will need the prime factorization of 48 at one step. So, you must know how to write a number as the product of its primes. Note that you can also use the discriminant method to get the answer but it will not be a much different process that we have done above. You will still need the prime factorization.
Complete step by step answer:
Here, we have been provided with the quadratic expression: - \[{{x}^{2}}-48=0\] and we are asked to solve it. That means we have to find the values of x.
Now, as we can see that the given equation is quadratic in nature, so we must have two roots or values of x. Since, the coefficient of x in the above equation is 0, so we will not apply the middle term split method. Here, let us apply the factorization method using the algebraic identity: - \[{{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)\].
Now, writing 48 as the product of its primes, we get,
\[\Rightarrow 48=2\times 2\times 2\times 2\times 3\]
This can be written as: -
\[\begin{align}
& \Rightarrow 48={{2}^{2}}\times {{2}^{2}}\times 3 \\
& \Rightarrow 48={{\left( 2\times 2 \right)}^{2}}\times 3 \\
& \Rightarrow 48={{4}^{2}}\times 3 \\
\end{align}\]
Here, we need to convert the factor 3 such that its exponent becomes 2 and we can apply the algebraic identity. To do this we will take the square root of 3 and to balance we will square it. So, we have,
\[\begin{align}
& \Rightarrow 48={{4}^{2}}\times {{\left( \sqrt{3} \right)}^{2}} \\
& \Rightarrow 48={{\left( 4\sqrt{3} \right)}^{2}} \\
\end{align}\]
Substituting this converted form of 48 in the quadratic equation: -
\[{{x}^{2}}-48\], we get,
\[\Rightarrow {{x}^{2}}-{{\left( 4\sqrt{3} \right)}^{2}}=0\]
Using the algebraic identity: - \[{{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)\] we get,
\[\Rightarrow \left( x+4\sqrt{3} \right)\left( x-4\sqrt{3} \right)=0\]
Substituting each term equal to 0, we get,
\[\Rightarrow \left( x+4\sqrt{3} \right)=0\] or \[\left( x-4\sqrt{3} \right)=0\]
\[\Rightarrow x=-4\sqrt{3}\] or \[x=4\sqrt{3}\]
Hence, the roots of the given quadratic equation are: - \[4\sqrt{3}\] and \[-4\sqrt{3}\].
Note:
One may note that whatever method we may apply we will need the prime factorization of 48 at one step. So, you must know how to write a number as the product of its primes. Note that you can also use the discriminant method to get the answer but it will not be a much different process that we have done above. You will still need the prime factorization.
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