Question

If zeros of ax2+bx+c be in ratio 4:5 show that 20b2=81ac

Answer 1

Hello mate,

Here is your answer,

Let the zeroes be 4α and 5α.

4α + 5α =  -b/a

9α = -b/a

α=-b/9a

c/a = 4α . 5α

c/a = 20α2

c/a = 20(-b/2a)2

c/a = 20b2/81a2

81a2c = 20ab2

81ac=20b2

Answer 2

Answer:

$20b^{2} =81ac$

Step-by-step explanation:

A polynomial function is a function that involves only non-negative integer powers or only positive integer exponents of a variable in an equation like the quadratic equation, cubic equation, etc.

The zeros of a polynomial p(x) are all the x-values that make the polynomial equal to zero. They are interesting to us for many reasons, one of which is that they tell us about the x-intercepts of the polynomial's graph.

Zeroes of Polynomial are the real values of the variable for which the value of the polynomial becomes zero. So, real numbers, 'm' and 'n' are zeroes of polynomial p(x), if p(m) = 0 and p(n) = 0

Given: $ax^{2} +bx+c=0$ , zeros in ratio 4:5

Find: Show that $20b^{2} =81ac$

For given polynomial $ax^{2} +bx+c=0$,

Sum of roots$=\frac{b}{a}$

Product of roots$=\frac{c}{a}$

Let $\alpha$ and $\beta$ be the roots or say zeros of given polynomial.

So,

$\alpha :\beta =4:5$

$\alpha =4k,\beta =5k$

$\alpha +\beta =-\frac{b}{a} \\\Rightarrow 4k+5k=-\frac{b}{a}\\\Rightarrow 9k=-\frac{b}{a}\\\Rightarrow k=-\frac{b}{9a}$

Product of roots:

$\alpha \beta =\frac{c}{a} \\\Rightarrow 4k\times5k= =\frac{c}{a}\\\Rightarrow 20k^{2} =\frac{c}{a}$

Substitute value of k in above equation:

$\Rightarrow 20\left ( \frac{-b}{9a} \right )^{2}=\frac{c}{a}\\\Rightarrow 20\left ( \frac{b^{2} }{81a^{2}} \right )=\frac{c}{a}\\\Rightarrow 20b^{2} =81ac$

Hence, proved

#SPJ2