Question

Solve
\[3x - 8\sqrt x - 60 = 0\]

Answer 1

Hint: In the given problem we need to solve this for ‘x’. Here we need to remove the radical symbol \[\sqrt {} \]. To do that we substitute \[\sqrt x = t\] in the given equation and after simplifying we will have a quadratic equation. We can solve this using the factorization method. The most important step is, after obtaining the roots we need to replace the variable ‘t’ by ‘x.

Complete step-by-step answer:
Now we have,
\[3x - 8\sqrt x - 60 = 0\]
Put \[\sqrt x = t\], then we have \[x = {t^2}\]. Then the given equation becomes,
\[3{t^2} - 8t - 60 = 0\]
On comparing the given equation with the standard quadratic equation \[a{t^2} + bt + c = 0\]. We have \[a = 3\], \[b = - 8\] and \[c = - 60\].
For factorization, the standard equation is rewritten as \[a{x^2} + {b_1}x + {b_2}x + c = 0\] such that\[{b_1} \times {b_2} = ac\] and\[{b_1} + {b_2} = b\].
Here we can say that \[{b_1} = - 18\] and \[{b_2} = 10\]. Because \[{b_1} \times {b_2} = - 180\] \[(a \times c)\] and \[{b_1} + {b_2} = - 8(b)\].
Now we write \[3{t^2} - 8t - 60 = 0\] as,
\[3{t^2} - 18t + 10t - 60 = 0\]
Taking ‘3t’ common in the first two terms and taking 10 common in the remaining two terms we have,
\[3t(t - 6) + 10(t - 6) = 0\]
Again taking \[(t - 6)\] common we have,
\[(t - 6)(3t + 10) = 0\]
By zero product principle we have
\[ \Rightarrow t - 6 = 0\] and \[3t + 10 = 0\]
\[ \Rightarrow t = 6\] and \[3t = - 10\]
\[ \Rightarrow t = 6\] and \[t = \dfrac{{ - 10}}{3}\].
But we have substituted \[\sqrt x = t\] and if the value of ‘x’ is negative the root will be a complex number. So we neglect \[t = \dfrac{{ - 10}}{3}\]
Then \[t = 6\] and \[\sqrt x = 6\]
Squaring on both sides we have,
\[ \Rightarrow x = 36\] is the required answer.

Note: The highest exponent of the polynomial in a polynomial equation is called its degree. A polynomial equation has exactly as many roots as its degree. In various fields of mathematics require the point at which the value of a polynomial is zero, those values are called the factors/solution/zeros of the given polynomial. On the x-axis, the value of y is zero so the roots of an equation are the points on the x-axis, that is the roots are simply the x-intercepts.