Question

if alpha and beta are the zeroes of the polynomial p(x)=x²-5x+6,find the value of alpha⁴beta²+alpha²beta⁴​

Answer 1

ANSWER:

• The value of above expression = 468.

GIVEN:

• P(x) = x²-5x+6

TO FIND:

$\alpha \: {}^{4} \beta {}^{2} + \alpha {}^{2} \beta {}^{4}$

SOLUTION:

P(x) = x²-5x+6

Here;

$\implies \alpha \: + \beta \: = \frac{ - ( - 5)}{1} \\ \\ \implies \: \alpha \: \times \beta = \frac{6}{1}$

Now Simplify the above expression

$= \alpha \: {}^{4} \beta {}^{2} + \alpha {}^{2} \beta {}^{4} \\ = \alpha {}^{2} \beta {}^{2} ( \alpha {}^{2} + \beta {}^{2} ) \: \: \: \: ......(i)$

Now finding the value of (α² + β²)

$= \alpha {}^{2} + \beta {}^{2} \\ = ( \alpha \: + \beta ){}^{2} - 2 \alpha \beta \\ \\ putting \: the \: values \: we \: get \\ = (5) {}^{2} - 2 \times 6 \\ = 25 - 12 \\ = 13$

Now;

putting (α² + β²)= 13 in eq(i) we get:

$=( \alpha \beta) {}^{2} (13) \\ = (6) {}^{2} \times 13 \\ = 36 \times 13 \\ = 468$

The value of above expression = 468.