Question

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

Answer 1

Given:

We have given a quadratic equation$x^2-7x-18=0$.

To Find:

We have find the diffence of the root?

Step-by-step explanation:

• Since , an quadratic equation is given to us which is $x^2-7x-18=0$

• Roots of quadratic equation are also known as the zeroes of the equation.

• We always have two roots of any quadratic equation because in the quadratic equation the heigest degree of the equation is two.

        We have $x^2-7x-18=0$

• First of all we will factorize the given quadratic equation as follows:

       $x^2-7x-18=0\\x^2-9x+2x-18=0$

• Now we will take the coomon terms out and write as

       $x(x-9)+2(x-9)=0\\\\(x+2)(x-9)=0\\x-9=0,x+2=0\\x=9,x=-2$

• Hence the roots of equation are 9,-2

• Now the difference of roots is 9-(-2)=11

Hence, answer is 11.

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 $11$.