Question

a and b are the zeros of the polynomial x^2+3x-8,can you find the value of a^2+b^2,a^3+b^3 and a^4+b^4.​

Answer 1

EXPLANATION.

α,β are the zeroes of the polynomial,

⇒ F(x) = x² + 3x - 8.

As we know that,

Sum of zeroes of quadratic equation,

⇒ α + β = -b/a.

⇒ α + β = -3/1 = -3.

Products of zeroes of quadratic equation,

⇒ αβ = c/a.

⇒ αβ = -8.

To Find value of :

(1) = α² + β².

As we know that,

Formula of : a² + b² = (a + b)² - 2ab.

⇒ α² + β² = (α + β)² - 2αβ.

⇒ α² + β² = (-3)² - 2(-8).

⇒ α² + β² = 9 + 16.

⇒ α² + β² = 25.

(2) = α³ + β³.

As we know that,

Formula of : a³ + b³ = (a + b)(a² - ab + b²).

⇒ α³ + β³ = (α + β)(α² - αβ + β²).

⇒ α³ + β³ = (α + β)(α² + β² - αβ).

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

⇒ α³ + β³ = (α + β)[(α + β)² - 3αβ].

⇒ α³ + β³ = (-3)[(-3)² - 3(-8)].

⇒ α³ + β³ = -3[9 + 24].

⇒ α³ + β³ = -3[33].

⇒ α³ + β³ = - 99.

(3) = α⁴ + β⁴.

As we know that,

Formula of : a⁴ + b⁴ = (a² + b²)² - 2a²b².

⇒ α⁴ + β⁴ = (α² + β²)² - 2α²β².

⇒ α⁴ + β⁴ = [(α + β)² - 2αβ]² - 2(αβ)².

⇒ α⁴ + β⁴ = [(-3)² - 2(-8)]² - 2(-8)².

⇒ α⁴ + β⁴ = [9 + 16]² - 2(64).

⇒ α⁴ + β⁴ = [25]² - 128.

⇒ α⁴ + β⁴ = 625 - 128.

⇒ α⁴ + β⁴ = 497.

Answer 2

${\large{\bold{\rm{\underline{Given \: that}}}}}$

${\small{\sf{\bigstar \alpha \: and \: \beta \: are \: zeros \: of \: polynomial \: \bf \: x^{2}+3x-8}}}$

${\large{\bold{\rm{\underline{To \: find}}}}}$

${\small{\sf{\bigstar \alpha^{2} + \beta^{2}}}}$

${\small{\sf{\bigstar \alpha^{3} + \beta^{3}}}}$

${\small{\sf{\bigstar \alpha^{4} + \beta^{4}}}}$

${\large{\bold{\rm{\underline{Solution}}}}}$

${\small{\sf{\bigstar \alpha^{2} + \beta^{2} \: \bf \: 25}}}$

${\small{\sf{\bigstar \alpha^{3} + \beta^{3} \: \bf \: 99}}}$

${\small{\sf{\bigstar \alpha^{4} + \beta^{4} \: \bf \: 497}}}$

${\large{\bold{\rm{\underline{Using \: concept}}}}}$

${\small{\sf{\bigstar Sum \: of \: quadratic \: equation \: is \: given \: by \: what?}}}$

${\small{\sf{\bigstar Product \: of \: quadratic \: equation \: is \: given \: by \: what?}}}$

${\large{\bold{\rm{\underline{Using \: formulas}}}}}$

${\small{\sf{\bigstar Sum \: of \: quadratic \: equation \: is \: given \: by \: \bf \: \alpha + \beta \: = -b/a}}}$

${\small{\sf{\bigstar Product \: of \: quadratic \: equation \: is \: given \: by \: \bf \: \alpha \beta \: = c/a}}}$

${\small{\sf{\bigstar Some \: identities}}}$

${\large{\bold{\rm{\underline{Full \; Solution}}}}}$

~ As the given polynomial is ${\red{x^{2}+3x-8}}$ and ${\red{\alpha}}$ and ${\red{\beta}}$ are it's zeros..!

$\; \; \; \; \; \; \; \; \; \;{\frak{\underline{\dag \: As \: we \: know \: that \: the}}}$

~ The sum of any quadratic polynomial is given by ${\red{\alpha + \beta \: = -b/a}}$

${\sf{:\implies \alpha + \beta \: = -b/a}}$

${\sf{:\implies \alpha + \beta \: = -3/1}}$

${\sf{:\implies \alpha + \beta \: = -3}}$

~ The sum of any quadratic polynomial is given by ${\red{\alpha \beta \: = -c/a}}$

${\sf{:\implies \alpha \beta \: = -c/a}}$

${\sf{:\implies \alpha \beta \: = -8}}$

~ Let's find ${\small{\sf{\alpha^{2} + \beta^{2}}}}$

${\small{\boxed{\bf{Using \: identity \: = a^{2} + b^{2} \: = (a+b)^{2} \: -2ab}}}}$

${\sf{:\implies \alpha^{2} + \beta^{2} \: = (\alpha \beta)^{2} \: -2 \alpha \beta}}$

${\sf{:\implies \alpha^{2} + \beta^{2} \: = (-3)^{2} -2(-8)}}$

${\sf{:\implies \alpha^{2} + \beta^{2} \: = (-3 \times -3) \: -2(-8)}}$

${\sf{:\implies \alpha^{2} + \beta^{2} \: = (3 \times 3) \: -2(-8)}}$

${\sf{:\implies \alpha^{2} + \beta^{2} \: = 9 \: -2(-8)}}$

${\sf{:\implies \alpha^{2} + \beta^{2} \: = 9 \: (-2 \times -8)}}$

${\sf{:\implies \alpha^{2} + \beta^{2} \: = 9 + 16}}$

${\sf{:\implies \alpha^{2} + \beta^{2} \: = 25}}$

~ Now let's find ${\small{\sf{\alpha^{3} + \beta^{3}}}}$

${\small{\boxed{\bf{Using \: identity \: = a^{3} + b^{3} \: = (a+b)(a^{2} -ab + b^{2}}}}}$

${\sf{:\implies \alpha^{3} + \beta^{3} \: = (\alpha + \beta)(\alpha^{2} -\alpha \beta + \beta^{2}}}$

${\sf{:\implies \alpha^{3} + \beta^{3} \: = (\alpha + \beta)(\alpha^{2} + \beta^{2} - \alpha \beta}}$

${\sf{:\implies \alpha^{3} + \beta^{3} \: = (\alpha + \beta)[(\alpha + \beta)^{2} -2 \alpha \beta - \alpha \beta}}$

${\sf{:\implies \alpha^{3} + \beta^{3} \: = (\alpha + \beta)[(\alpha + \beta)^{2} -3 \alpha \beta}}$

${\sf{:\implies \alpha^{3} + \beta^{3} \: = (-3)[(-3)^{2} -3(-8)}}$

${\sf{:\implies \alpha^{3} + \beta^{3} \: = (-3)(-3 \times -3) \: -3(-8)}}$

${\sf{:\implies \alpha^{3} + \beta^{3} \: = (-3)(3 \times 3) \: -3(-8)}}$

${\sf{:\implies \alpha^{3} + \beta^{3} \: = (-3) 9 \: -3(-8)}}$

${\sf{:\implies \alpha^{3} + \beta^{3} \: = (-3) 9 +24}}$

${\sf{:\implies \alpha^{3} + \beta^{3} \: = (-3) 33}}$

${\sf{:\implies \alpha^{3} + \beta^{3} \: = -99}}$

~ Now let's find ${\small{\sf{\alpha^{4} + \beta^{4}}}$

${\small{\boxed{\bf{Using \: identity \: = a^{4} + b^{4} \: = (a^{2}+b^{2})^{2} -2a^{2}b^{2}}}}}$

${\sf{:\implies \alpha^{4} + \beta^{4} \: = (\alpha^{2}+\beta^{2})^{2} -2\alpha^{2}\beta^{2}}}$

${\sf{:\implies \alpha^{4} + \beta^{4} \: = [(\alpha + \beta)^{2} - \alpha \beta]^{2} \: - 2(\alpha \beta)^{2}}}$

${\sf{:\implies \alpha^{4} + \beta^{4} \: = [(-3)^{2} -2(-8)]^{2} -2(-8)^{2}}}$

${\sf{:\implies \alpha^{4} + \beta^{4} \: = [9 + 16]^{2} -2(64)}}$

${\sf{:\implies \alpha^{4} + \beta^{4} \: = (-25)^{2} -2(64)}}$

${\sf{:\implies \alpha^{4} + \beta^{4} \: = 625 -2(64)}}$

${\sf{:\implies \alpha^{4} + \beta^{4} \: = 625 - 128}}$

${\sf{:\implies \alpha^{4} + \beta^{4} \: = 497}}$