Question

If alpha and beta zeroes of 3x²-x-4 find the value of alpha⁴beta³ + alpha³beta⁴.

[Ans: -64/81]​

Answer 1

EXPLANATION.

α,β are the zeroes of the Quadratic Equation,

⇒ p(x) = 3x² - x - 4.

As we know that,

Sum of zeroes of the quadratic equation,

⇒ α + β = -b/a.

⇒ α + β = -(-1)/3 = 1/3.

Products of zeroes of the quadratic equation,

⇒ αβ = c/a.

⇒ αβ = -4/3.

To find value of = (α⁴β³ + α³β⁴).

⇒ (α⁴β³ + α³β⁴).

⇒ α³β³(α + β).

⇒ (αβ)³(α + β).

⇒ (-4/3)³(1/3).

⇒ (-64/27).(1/3).

⇒ (-64/81).

Value of (α⁴β³ + α³β⁴) = -64/81.

                                                                                                                                       

MORE INFORMATION.

Location of Roots of a quadratic Equation ax² + bx + c = 0.

(A) = Conditions for both the roots will be greater than k.

(1) = D ≥ 0.

(2) = k < -b/2a.

(3) = af(k) > 0.

(B) = Conditions for both the roots will be less than k.

(1) = D ≥ 0.

(2) = k > -b/2a.

(3) = af(k) > 0.

(C) = Conditions for k lie between the roots.

(1) = D > 0.

(2) = af(k) < 0.

(D) = Conditions for exactly one roots lie in the interval (k₁, k₂) where k₁ < k₂.

(1) = f(k₁)f(k₂) < 0.

(2) = D > 0.

(E) = When both roots lie in the interval (k₁, k₂) where k₁ < k₂.

(1) = D > 0.

(2) = f(k₁).f(k₂) > 0.

(F) = Any algebraic expression f(x) = 0 in interval [a, b] if,

(1) = sign of f(a) and f(b) are of same than either no roots or even no. of roots exists.

(2) = sign of f(a) and f(b) are opposite then f(x) = 0 has at least one real roots or odd no. of roots.

                                           

Answer 2

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

✠ If alpha and beta are zeroes of 3x²-x-4 ; find the value of alpha⁴beta³ + alpha³beta⁴.

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

★ α (alpha) and β (beta) are zeroes of quadratic equation = 3x²-x-4

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

★ The value of α⁴β³ + α³β⁴

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

★ The value of α⁴β³ + α³β⁴ = ${\sf{\bold{\red{-64/81}}}}$

${\large{\bold{\rm{\underline{Using \; concepts}}}}}$

★ Sum of zeros of any quadratic equation is given by what ?

★ Product of zeros of any quadratic equation is given by what ?

${\large{\bold{\rm{\underline{Using \; dimensions}}}}}$

★ Sum of zeros of any quadratic equation is given by ➝ α+β = -b/a

★ Product of zeros of any quadratic equation is given by ➝ αβ = c/a

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

~ Let us take a view on the sum of zeros of any quadratic equation

⇢ α+β = -b/a

⇢ α+β = -(-1)/3

• - and - cancel each other !..

⇢ α+β = 1/3

~ Let us take a view on the product of zeros of any quadratic equation

⇢ αβ = c/a

⇢ αβ = -4/3

~ Now using the above information let's find out the the value of α⁴β³ + α³β⁴.

⇢ α⁴β³ + α³β⁴.

⇢ α³β³ ( α+β )

• Law of exponents !..

⇢ (αβ)³ ( α+β )

⇢ (-4/3)³(1/3)

⇢ -4/3 × -4/3 × -4/3 (1/3)

• - × - = + always !..

⇢ 16/9 × -4/3 (1/3)

⇢ (-64/27) (1/3)

⇢ (-64/27) × (1/3)

⇢ -64/27 × 1/3

⇢ -64/81

${\frak{Henceforth, \: -64/81 \: is \: the \: value \: of \: \alpha^{4} \beta^{3} + \alpha^{3} \beta^{4}}}$

${\large{\bold{\rm{\underline{Explore \; more}}}}}$

Knowledge about Quadratic equations -

★ Sum of zeros of any quadratic equation is given by ➝ α+β = -b/a

★ Product of zeros of any quadratic equation is given by ➝ αβ = c/a

★ A quadratic equation have 2 roots

★ ax² + bx + c = 0 is the general form of quadratic equation

Law of Exponents -

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto a^{m} \times a^{n} = \: a^{m+n}}}}$

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto (a^{m})^{n} \: = a^{mn}}}}$

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto a^{m} \times b^{m} = (ab)^{m}}}}$

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto a^{0} = 1}}}$

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto \dfrac{a^{m}}{b^{m}} = (\dfrac{a}{b})^{m}}}}$

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto \dfrac{a^{m}}{a^{n}} = a^{m-n}}}}$

Where, m - n ∈ N

Easy to remember about last rule ⬆️

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto x^{a} ÷ x^{b} = x^{a-b}}}}$

It happens when a > b

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto x^{a} ÷ x^{b} = \dfrac{1}{x^{b-a}}}}}$

It's happen when a < b

Law of Exponents -

$\begin{gathered}\begin{gathered}\boxed{\begin{minipage}{5 cm}\bf{\dag}\:\:\underline{\text{Law of Exponents :}}\\\\\bigstar\:\:\tt\dfrac{a^m}{a^n} = a^{m - n}\\\\\bigstar\:\:\tt{(a^m)^n = a^{mn}}\\\\\bigstar\:\:\tt(a^m)(a^n) = a^{m + n}\\\\\bigstar\:\:\tt\dfrac{1}{a^n} = a^{-n}\\\\\bigstar\:\:\tt\sqrt[\tt n]{\tt a} = (a)^{\dfrac{1}{n}}\end{minipage}}\end{gathered}\end{gathered}$