Answer :
Answer :
- x = 3 or x = -13
⠀
Explanation :
[tex]\longrightarrow \sf \qquad {x}^{2} + 10x - 39 = 0[/tex]
⠀
We have to find the two numbers a and b such that,
[tex]\longrightarrow \sf \qquad a + b = 10[/tex]
[tex]\longrightarrow \sf \qquad a b = 39[/tex]
⠀
Obviously, the two numbers are 3 and 13.
[tex]\longrightarrow \sf \qquad {x}^{2} - 3x + 13x - 39 = 0[/tex]
[tex]\longrightarrow \sf \qquad {x}(x - 3)+ 13(x - 3) = 0[/tex]
[tex]\longrightarrow \sf \qquad ({x}+ 13)(x - 3) = 0[/tex]
⠀
Whether, the value of x :
[tex]\longrightarrow \sf \qquad {x}+ 13 = 0[/tex]
[tex]\longrightarrow \pmb{\bf \qquad {x} = - 13}[/tex]
⠀
Whether, the value of x :
[tex]\longrightarrow \sf \qquad {x} - 3 = 0[/tex]
[tex]{\longrightarrow { \pmb{\bf \qquad {x} = 3}}}[/tex]
⠀
We are given with the equation x² + 10x - 39 = 0 and need to find x, so let's start ;
[tex]{:\implies \quad \sf x^{2}+10x-39=0}[/tex]
By using splitting the middle term method, Rewrite as ;
[tex]{:\implies \quad \sf x^{2}+13x-3x-39=0}[/tex]
[tex]{:\implies \quad \sf x(x+13)-3(x+13)=0}[/tex]
[tex]{:\implies \quad \sf (x+13)(x-3)=0}[/tex]
So, here either (x + 13) = 0 or (x - 3) = 0, when you equate both of them with 0, you will get x = -13 and x = 3
Hence, The required answer is -13 and 3