Question

find the numerical difference of the roots of equation x2-7x-18=0​

Answer 1

Answer:

Here,

x² - 7x - 18 = 0

0 has no value

So,

x² - 7x - 18

Here we have to find just root difference of

x² - 7x - 18

=x² + 2x - 9x - 18

= x(x + 2) -9 (x + 2)

= (x+2)(x-9)

So,

Root difference is

So the answer is above

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Answer 2

Given a quadratic equation x^2-7x-18=0, find the difference between its root.

Explanation:

Given a quadratic equation of the form ax^2+bx+c=0 where a, b and c are real numbers having roots \alpha, \beta is given by,                                                    \alpha=\frac{-b+\sqrt{b^2-4ac} }{2a} \ \ \ \ \ \ \ \ \ \ \ \ \beta=\frac{-b-\sqrt{b^2-4ac} }{2a}  

Hence for the difference of roots \alpha-\beta we get,                                                 \alpha-\beta=\frac{-b+\sqrt{b^2-4ac} }{2a} -\frac{-b-\sqrt{b^2-4ac} }{2a}\\->\alpha-\beta=\frac{\sqrt{b^2-4ac} }{a}   -----(a)

now we have the equation x^2-7x-18=0 where, a=1,\ b=-7\ and\ c=-18  

hence from (a) we get the difference of the roots as,                                         \alpha-\beta=\sqrt{(-7)^2-4(1)(-18)} \\->\alpha-\beta=\sqrt{49+72}\\->\alpha-\beta=\sqrt{121}\\->\alpha-\beta=11  

The numerical difference of the roots is 11.