Question

if alpha and beta are the zeroes of 3x^2+2x+3 then find 1 by alpha^2+1 by beta^2

Answer 1

Answer:

-5/9

Step-by-step explanation:

Let us consider the roots of the equation

3x² + 2x + 3 be

$\alpha \: and \: \: \beta$

So we know that

Sum of the roots of a quadratic equation is equal to the minus of co-efficient of x divided by co-efficient of x²

$\alpha + \beta = \frac{ - 2}{3} \\ ({ \alpha + \beta) }^{2} = \frac{4}{9} \\ \implies { \alpha }^{2} + { \beta }^{2} + 2 \alpha \beta = \frac{4}{9}$

And also product of the roots is equal to constant term divided by co-efficient of x ²

$\alpha \beta = \frac{3}{3} \\ \implies \alpha \beta = 1 \\ \implies \alpha ^{2} \beta ^{2} = 1$

Again putting the value of product of roots in the previous equation

${ \alpha }^{2} + \beta ^{2} + 2 = \frac{4}{9} \\ \implies \ { \alpha }^{2} + \beta ^{2} = \frac{4}{9} - 2 \\ \implies { \alpha }^{2} + \beta ^{2} = - \frac{5}{9}$

Now dividing it by square of product of roots

$\frac{ { \alpha }^{2} + { \beta }^{2} }{ { \alpha }^{2} \beta {}^{2} } = \frac{ - 5}{9} \\ \implies \frac{1}{ { \alpha }^{2} } + \frac{1}{ \beta {}^{2} } = \frac{ - 5}{9}$

Answer 2

Solutions:

Let the consider roots of the equation.

3x² + 2x + 3

Sum of the roots of quadratic equations.

$\alpha + \beta = - 2 \div 3 \\ \alpha + \beta = 4 \div 9 \\ \alpha {}^{2} + \beta {}^{2} + 2 \alpha \beta = 4 \div 9 \\ \alpha \beta = 3 \div 3 \\ \alpha \beta = 1 \\ \alpha {}^{2} \beta {}^{2} = 1$

$\alpha {}^{2} + \beta {}^{2} + 2 = 4 \div 9 \\ \alpha {}^{2} \beta {}^{2} = 4 \div 9 - 2 \\ \alpha {}^{2} + \beta {}^{2} = - 5 \div 9$

$\alpha {}^{2} + \beta {}^{2} \div \alpha {}^{2} \beta {}^{2} = - 5 \div 9 \\ 1 \div \alpha {}^{2} + 1 \div \beta {}^{2} = - 5 \div 9$

silentlover45.❤️