Question

If α and β are the zeroes of the polynomial x2

– 7x + 10, then find the value of α3 + β3​

Answer 1

Answer:

133

Step-by-step explanation:

${x}^{2} - 7x +1 0 = 0 \\ \\ \alpha + \beta = 7 \\ \\ \alpha \beta = 10 \\ \\ \\ \\ { \alpha }^{2} + { \beta }^{2} = {( \alpha + \beta )}^{2} - 2 \alpha \beta = {7}^{2} - 2 \times 10 = 49 - 20 = 29\\ { \alpha }^{3} + { \beta }^{3} = ( \alpha + \beta )( { \alpha }^{2} + { \beta }^{2} - \alpha \beta ) \\ \\ = 7 \times (29 - 10) \\ \\ = 7 \times 19 \\ \\ = 133$

Answer 2

EXPLANATION.

α and β are the zeroes of the equation.

⇒ p(x) = x² - 7x + 10.

As  we know that,

Sum of the zeroes of the quadratic equation.

⇒ α + β = -b/a.

⇒ α + β = -(-7)/1 = 7.

Products of the zeroes of the quadratic equation.

⇒ αβ = c/a.

⇒ αβ = 10.

To find :

⇒ α³ + β³.

⇒ α³ + β³ = (α + β)[α² - αβ + β²].

⇒ α³ + β³ = (α + β)[(α + β)² - 2αβ - αβ].

⇒ α³ + β³ = (α + β)[(α + β)² - 3αβ].

Put the values in the equation, we get.

⇒ α³ + β³ = (7)[(7)² - 3(10)].

⇒ α³ + β³ = (7)[49 - 30].

⇒ α³ + β³ = (7)[19].

⇒ α³ + β³ = 133.

                                                                                                                   

MORE INFORMATION.

Nature of the roots of the quadratic expression.

(1) = Real and unequal, if b² - 4ac > 0.

(2) = Rational and different, if b² - 4ac is a perfect square.

(3) = Real and equal, if b² - 4ac = 0.

(4) = If D < 0 Roots are imaginary and unequal Or complex conjugate.