Question

if the quadratic equation x^2+6x-9=0 has two real roots m and n, what is 1/m+1/n?
I'm kinda confused I did the quadratic formula and got 3+3√2, so what would the roots be

Answer 1

Answer:

$\qquad\qquad\boxed{\red{\sf \dfrac{1}{m}+\dfrac{1}{n}=\dfrac{2}{3} }}$

Step-by-step explanation:

Given that , the quadratic equation has two real roots , m and n . We need to find the value of 1/m + 1/n . The given equation is ,

$\sf\red{\dashrightarrow}\: x^2 +6x -9 = 0$

• We can find out this value by finding the sum of roots and the roots of roots. You used quadratic formula to find out the roots but here we can directly use the results of Sum of Roots and product of roots.

• The standard form of a quadratic equation is ax² + bx + c = 0. With respect to Standard form we have the sum of roots as -b/a . And the product of roots as c/a .

• Here , wrt to Standard form , we have , a = 1 , b = 6 and c = (-9) . Now simplify the given equation in the form of sum and product of roots whose value is to be finded out.

$\sf\dashrightarrow \dfrac{1}{m}+\dfrac{1}{n} \\\\\sf\dashrightarrow \dfrac{ m + n}{mn}$

• Here we will notice that , the numerator is in the form of sum of roots and the denominator is in the form of product of roots. Substituting the respective values , we have ,

$\sf\dashrightarrow \dfrac{ m + n}{mn} \\\\\sf\dashrightarrow \dfrac{ \dfrac{-b}{a}}{\dfrac{c}{a}}\\\\\sf\dashrightarrow \dfrac{-b}{c}\\\\ \sf\dashrightarrow \boxed{\pink{\sf \dfrac{-6}{-9}=\dfrac{2}{3}}}$