Question

Solve \[(x - 3)(x - 4) = \dfrac{{34}}{{{{33}^2}}}\]

Answer 1

Hint: When we have a polynomial of the form \[a{x^2} + bx + c = 0\], we can solve this quadratic with splitting up the b term into two terms based on the signs of a and c terms. If we have the opposite sign for both a and c terms, then we split the b term into two parts. These parts are formed such that the difference of them is b term itself and their product is the same as that of the product of a and c terms.

Complete step-by-step answer:
To find the roots from the given quadratic equation:
From the given \[(x - 3)(x - 4) = \dfrac{{34}}{{{{33}^2}}}\]
\[{33^2} = 1089\]
\[{x^2} - 4x - 3x + 12 = \dfrac{{34}}{{1089}}\]
\[{x^2} - 7x + 12 = \dfrac{{34}}{{1089}}\]
Cross multiplying we get,
\[1089{x^2} - 7623x + 13068 = 34\]
\[1089{x^2} - 7623x + 13034 = 0\]
In the given quadratic polynomial \[1089{x^2} - 7623x + 13034 = 0\], the coefficient of \[{x^2}\]
and constant terms are of the same sign and their product is \[\dfrac{{13034}}{{1089}}\]. That is we have a and c coefficients with same sign in the polynomial of form \[a{x^2} + bx + c = 0\]
Hence, we would split \[ - 7623\] , which is the coefficient of \[x\], in two parts, whose sum is \[\dfrac{{7623}}{{1089}}\] and product is \[\dfrac{{13034}}{{1089}}\] .
We now split
\[ - 7623x\] into \[ - 4389x\] and \[ - 3234x\]
\[ \Rightarrow 1089{x^2} - 3234x - 4389x + 13034 = 0\]
We take the terms common in the first 2 terms and last 2 terms.
\[33x(33x - 98) - 133(33x - 98) = 0\]
Here, we have
\[ \Rightarrow (33x - 133)(33x - 98) = 0\]
Now we equate the equation, then we get
\[33x - 133 = 0\] or \[33x - 98 = 0\]
\[\therefore x = \dfrac{{133}}{{33}}\] or \[x = \dfrac{{98}}{{33}}\] is the solution \[(x - 3)(x - 4) = \dfrac{{34}}{{{{33}^2}}}\]
So, the correct answer is “\[ x = \dfrac{{133}}{{33}}\] or \[x = \dfrac{{98}}{{33}}\]”.

Note: If sign of coefficient of \[{x^2}\] and constant terms are same, then we factor the polynomial by splitting up the b term into two terms such that sum of those parts is b term and product is same as that of product of a and c terms. Factoring by grouping will not always work. In such cases we better go with the quadratic formula.