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Question

Question: Solve the given quadratic equation \[{x^2} - 10x - 24?\]...

Solve the given quadratic equation x210x24?{x^2} - 10x - 24?

Explanation

Solution

By using the factorization method to obtain the factors of the given equationx2+yx+z{x^2} + yx + z. Here we can use the mid term splitting method to factorise this equation in which we are going to split the middle term and can have required two factors.
For equation x2+yx+z{x^2} + yx + z Here we have to split “y” in such a way that the split number say “a” and “b” have the relationship a×b=z(1)a \times b = z(1) and ab=ya - b = yfor the above equation.

Formulae Used:
For equation x2+yx+z{x^2} + yx + z

Complete step by step solution:
For the given equation x210x24{x^2} - 10x - 24

Using mid term splitting rule we are breaking “10” into “12 and 2” as these two numbers follows the rule of mid term splitting, and now on solving the equation we get,

x2(122)x24=0 x212x(2x)24=0 x212x+2x24=0 x(x12)+2(x12)=0 (x12)(x+2)=0 x=12,2 \Rightarrow {x^2} - \left( {12 - 2} \right)x - 24 = 0 \\\ \Rightarrow {x^2} - 12x - ( - 2x) - 24 = 0 \\\ \Rightarrow {x^2} - 12x + 2x - 24 = 0 \\\ \Rightarrow x(x - 12) + 2(x - 12) = 0 \\\ \Rightarrow (x - 12)(x + 2) = 0 \\\ \Rightarrow x = 12, - 2 \\\

Above are the two values of the given quadratic equation.

Here we have to split “y” in such a way that the split number say “a” and “b” have the relationship a×b=z(1)a \times b = z(1) and ab=ya - b = yfor the above equation.

Additional Information: Mid term split method is easy to find factor, but another method that is taking the highest common factor directly and obtaining the factors. This method works sometimes in some questions only mostly for 3 variable questions.

Note: In two variable questions it will work but make the question very complicated, so you have to be careful while using the highest common factor method. This method is fast and easy to use but is applicable to certain specific questions only. After getting the roots of the given equation you can satisfy by putting it in the equation.