Question

How do you solve \[32{{x}^{2}}-3x-14={{\left( 2x-1 \right)}^{2}}\]?

Answer 1

Hint: This question is from the topic of algebra. We will use the Sridharacharya’s rule to solve this question. In this question, we will first break the square term that is on the right side of the equation. After that, we will take all the variables and the constants to the left side of the equation. After that, we will add the summable terms. After doing further solving, we will find out the value of x using Sridharacharya’s rule.

Complete step by step solution:
Let us solve this question.
In this question we have to solve the term \[32{{x}^{2}}-3x-14={{\left( 2x-1 \right)}^{2}}\]. That means we have to find the value of x from the equation \[32{{x}^{2}}-3x-14={{\left( 2x-1 \right)}^{2}}\].
The equation we have solve is
\[32{{x}^{2}}-3x-14={{\left( 2x-1 \right)}^{2}}\]
As we know that the term \[{{\left( a-b \right)}^{2}}\] can also be written as \[{{a}^{2}}+{{b}^{2}}-2\times a\times b\]. So, using this, we can write the above equation as
\[\Rightarrow 32{{x}^{2}}-3x-14={{\left( 2x \right)}^{2}}+{{\left( 1 \right)}^{2}}-2\times 2x\times 1\]
The above equation can also be written as
\[\Rightarrow 32{{x}^{2}}-3x-14=4{{x}^{2}}+1-4x\]
Now, taking all the terms of \[{{x}^{2}}\], \[x\] and constants to the left side of equation, we get
\[\Rightarrow 32{{x}^{2}}-4{{x}^{2}}-3x+4x-14-1=0\]
The above equation can also be written as
\[\Rightarrow 28{{x}^{2}}+x-15=0\]
Now, we will find the value of x using Sridharacharya’s rule, we will get
\[x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}=\dfrac{-1\pm \sqrt{{{1}^{2}}-4\times 28\times 15}}{2\times 28}\]
The above can also be written as
\[\Rightarrow x=\dfrac{-1\pm \sqrt{1+1680}}{56}=\dfrac{-1\pm \sqrt{1681}}{56}\]
The above can also be written as
\[\Rightarrow x=\dfrac{-1\pm 41}{56}\]
From here, we get that the values of x are \[\dfrac{-1+41}{56}\] and \[\dfrac{-1-41}{56}\].
Hence, the values of x are \[\dfrac{40}{56}\] and \[\dfrac{-42}{56}\]
Therefore, we can say that the values of x are \[\dfrac{5}{7}\] and \[\dfrac{-6}{8}\].
Now, we have solved the term \[32{{x}^{2}}-3x-14={{\left( 2x-1 \right)}^{2}}\] and got the values of x as \[\dfrac{5}{7}\] and \[\dfrac{-6}{8}\].

Note: We should have a better knowledge in the topic of algebra and quadratic equation to solve this type of question easily. Always remember the formula that \[{{\left( a-b \right)}^{2}}\] is equal to \[{{a}^{2}}+{{b}^{2}}-2\times a\times b\].
Always remember that if we have given a general quadratic equation as \[a{{x}^{2}}+bx+c=0\].
Then according to Sridharacharya’s rule, the value of x will be
\[x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\].
Whenever we have to divide the fractions like \[\dfrac{40}{56}\] and \[\dfrac{-42}{56}\] & make them in the simple form, then first find out the prime factorization of all numbers that are numerator and denominator. After that, we will cancel out the same terms that are in numerator and denominator. Then, we will get the fractional term as simple term. We have got \[\dfrac{40}{56}\] and \[\dfrac{-42}{56}\] as \[\dfrac{5}{7}\] and \[\dfrac{-6}{8}\] in the above.