Question

Find the zeros of f(x) =x^2/a+b/ac*x/b-1/c​

Answer 1

Zeros of the equation are

$x_2=\dfrac{-b-\sqrt{b^2+2ac}}{2c} \\x_1=\dfrac{-b+\sqrt{b^2+2ac}}{2c}$

Explanation:

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Given that

$f(x)=\dfrac{x^2}{a}+\dfrac{b}{ac}x-\dfrac{1}c}$

f(x)=0 ,to find zeros

The above equation is a quadratic equation

we know that the general equation of quadratic

$Ax^2+Bx+C=0$

The zeros of the equation given as follows

$x_1=\dfrac{-B+\sqrt{B^2-2AC}}{2A}\\x_2=\dfrac{-B-\sqrt{B^2-2AC}}{2A}$

Now by putting the values in the above equation

$x_1=\dfrac{-\dfrac{b}{ac}+\sqrt{\dfrac{b^2}{a^2c^2}+2\times \dfrac{1}{a}\times \dfrac{1}{c}}}{2\times \dfrac{1}{a}}\\x_1=\dfrac{-b+\sqrt{b^2+2ac}}{2c}$

$x_2=\dfrac{-\dfrac{b}{ac}-\sqrt{\dfrac{b^2}{a^2c^2}+2\times \dfrac{1}{a}\times \dfrac{1}{c}}}{2\times \dfrac{1}{a}}\\x_2=\dfrac{-b-\sqrt{b^2+2ac}}{2c}$

Therefore the zeros of the equation will be

$x_2=\dfrac{-b-\sqrt{b^2+2ac}}{2c} \\x_1=\dfrac{-b+\sqrt{b^2+2ac}}{2c}$