Question

How do you solve \[x-15\sqrt{x}+56=0\]?

Answer 1

Hint: This question is from the topic of quadratic equation. In this question, we will find the value of x. In solving this question, we will convert the term x in the form of t that is \[x={{t}^{2}}\]. After that, we will solve the equation. We will find the value of t from the solved equation using Sridharacharya’s rule. After finding the value of t, we will square the values of t and then those terms will be the value of x.

Complete step by step solution:
Let us solve this question.
In this question, we have asked to solve the equation \[x-15\sqrt{x}+56=0\]. In this equation, we will find the value of x.
The given equation is
\[x-15\sqrt{x}+56=0\]
Now, let us take the term x as the square of t that is \[x={{t}^{2}}\]. Then, we can write the above equation as
\[\Rightarrow {{t}^{2}}-15\sqrt{{{t}^{2}}}+56=0\]
As we know that \[\sqrt{{{t}^{2}}}\] can also be written as \[t\], so the above equation can also be written as
\[\Rightarrow {{t}^{2}}-15t+56=0\]
Now, let us solve the above equation using Sridharacharya’s rule.
The value of t for the equation \[{{t}^{2}}-15t+56=0\] will be
\[t=\dfrac{-\left( -15 \right)\pm \sqrt{{{\left( -15 \right)}^{2}}-4\times 1\times 56}}{2\times 1}=\dfrac{15\pm \sqrt{225-224}}{2}\]
The above can also be written as
\[t=\dfrac{15\pm \sqrt{1}}{2}=\dfrac{15\pm 1}{2}\]
From here, we can say that the values of t are \[\dfrac{15+1}{2}\] and \[\dfrac{15-1}{2}\]
Or, we can say that the values of t are \[\dfrac{16}{2}\] and \[\dfrac{14}{2}\]
Hence, we get the values of t as 8 and 7.
Now, we know that we have taken x as \[{{t}^{2}}\].
So, we can say that the values of x will be \[{{8}^{2}}\] and \[{{7}^{2}}\] or we can say 64 and 49.

Hence, we have solved the equation \[x-15\sqrt{x}+56=0\] and have found the value of x as 64 and 49.

Note: We should have a better knowledge in the topic of quadratic equations. We should know about Sridharacharya’s rule. The Sridharacharya’s rule says that if we have given equation as \[a{{t}^{2}}+bt+c=0\], then the value of t from this equation will be \[t=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\]. We should know the square of some specific numbers like the square of 7 is 49 and the square of 8 is 64. And also remember that, the square root of a term will be equal to the same term or we can say \[\sqrt{{{t}^{2}}}=t\].