Question

If a and B are the zeroes of the polynomial

x square + x-6 find the value of 1 upon asquare + 1 upon b square​

Answer 1

EXPLANATION.

α,β are the zeroes of the polynomial,

⇒ p(x) = x² + x - 6.

To find value of ⇒ 1/α² + 1/β².

Sum of zeroes of quadratic polynomial,

⇒ α + β = -b/a.

⇒ α + β = -1.

Product of zeroes of quadratic polynomial,

⇒ αβ = c/a.

⇒ αβ = -6.

⇒ 1/α² + 1/β².

⇒ β² + α²/α²β².

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

⇒ (α + β )² - 2αβ/(αβ)².

⇒ (-1)² - 2(-6)/(-6)².

⇒ 1 + 12 / 36.

⇒ 13/36 = answer.

                                                       

MORE INFORMATION.

Location of roots of a quadratic equation,

ax² + bx + c.

(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) = D > 0.

(2) = f(k₁).f(k₂) > 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 then 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 root or odd no. of roots.

Answer 2

Step-by-step explanation:

______________________________

$\text{\Large\underline{\red{Question:-}}}$

$\Longrightarrow$ ● if alpha and beta are zeroes of the polynomial x² + x - 6, find the value of 1 / alpha²  + 1 / beta² .

$\text{\Large\underline{\orange{To\ find:-}}}$

● In this question we have to find the value of 1/α² + 1/β².

$\text{\Large\underline{\green{Given:-}}}$

● α and β are the zeroes of the polynomial x² + x - 6.

$\text{\Large\underline{\purple{Solution:-}}}$

● α and β are the zeroes of the polynomial:-

$\Longrightarrow$ $\sf{So,\ α\ +\ β\ =\ -1\ and\ αβ\ =\ -6.}$

Now:-

$\Longrightarrow$ $\sf{\dfrac{1}{α²}\ +\ \dfrac{1}{β²}\ =\ \dfrac{α²\ +\ β²}{α²β²}}$

$\Longrightarrow$ $\sf{\dfrac{(α²\ +\ β²)\ -\ 2αβ}{α²β²}}$

$\Longrightarrow$ $\sf{\dfrac{(-1)²\ -\ 2\ (-6)}{(-6)²}}$

$\Longrightarrow$ $\sf{\dfrac{1\ +\ 12}{36}}$

$\Longrightarrow$ $\sf\pink{\dfrac{13}{36}}$

Hence verified ✔

______________________________