Question

If roots of the equation x²+ ax + 25 = 0 are in the ratio of 2 : 3 then the value of a is​

Answer 1

Answer:

$x^2 + ax + 25=0$ is a quadratic equation of form $ax^2 + bx + c =0$.

The product of the roots is equal to $\frac{c}{a}=\frac{25}{1}=25$.

Let the roots be 2r and 3r.

$(2r)(3r)=25$

$6r^2 = 25$

$r^2 = \frac{25}{6}$

∴ $r=\sqrt\frac{25}{6}=\frac{5}{\sqrt{6}}=\frac{5\sqrt{6}}{6}$

∴ The roots are

$2r=\frac{5\sqrt{6}}{6}\times2=\frac{5\sqrt{6}}{3}$

$3r=\frac{5\sqrt{6}}{6}\times3=\frac{5\sqrt{6}}{2}$

Since the sum of the roots is equal to $\frac{-b}{a}$,

$\frac{5\sqrt{6}}{3}+\frac{5\sqrt{6}}{2}=-a$

Let $5\sqrt6$ be y.

$-a=\frac{2y+3y}{6}=\frac{5y}{6}$

$-a=\frac{25\sqrt{6}}{6}$

∴ $a=\frac{-25\sqrt{6}}{6}$

Answer: a = -25√6/6

I hope this answer helps. :D

Answer 2

Soln :-

a = -25√6

expln:-

Quadratic equations is in the form of,

x²-(α+β)x+αβ = 0

f(x) = x²+ax+25 = 0

by the equ,

α+β = -a, αβ = 25

ratio of the zeroes = 2 : 3 (given)

α = 2y, β = 3y

αβ = 25

(2y) (3y) = 25

6y² = 25

y² = 25/6

y = 5/√6

α = 2(5/√6) = 5√2/√3

β = 3(5/√6) = 5√3/√2

α+β = 5√2/√3 + 5√3/√2 = -a

(10+15) / √6 = -a

25/√6 = -a

a = -25/√6