Question

If a and ß are zeroes of
polynomial 2x2-3x+5, find
value of a/B + B/a​

Answer 1

EXPLANATION.

α and β are the zeroes of the polynomial.

⇒ 2x² - 3x + 5.

As we know that,

Sum of the zeroes of the quadratic equation.

⇒ α + β = -b/a.

⇒ α + β = -(-3)/2 = 3/2.

Products of the zeroes of the quadratic equation.

⇒ αβ = c/a.

⇒ αβ = 5/2.

To find :

⇒ α/β + β/α.

α² + β²/αβ.

As we know that,

Formula of :

⇒ (x² + y²) = (x + y)² - 2xy.

Using this formula in equation, we get.

⇒ [(α + β)² - 2αβ]/αβ.

⇒ [(3/2)² - 2(5/2)]/(5/2).

⇒ [9/4 - 5]/(5/2).

⇒ [9 - 20/4]/(5/2).

⇒ [-11/4/5/2].

⇒ -11/4 x 2/5.

⇒ -11/10.

⇒ α/β + β/α = -11/10.

                                                                                                                       

MORE INFORMATION.

Conjugate roots.

(1) = If D < 0.

One roots = α + iβ.

Other roots = α - iβ.

(2) = If D > 0.

One roots = α + √β.

Other roots = α - √β.

Answer 2

$\large\bf\underline\red{Question \: :- }$

If a and ß are zeroes of polynomial 2x2-3x+5, find

value of a/B + B/a ?

$\large\bf\underline\red{Answer \: :- }$

We know that,

a and ß are the zeros of 2x² - 3x + 5 polynomial.

We have to find :

a/ß + ß/a

First finding sum of the zeros

$\longmapsto\sf{ \alpha + \beta = \frac{ -b}{ a } } \: \: \: \: \: \: \: \\ \\ \longmapsto\sf{ \alpha + \beta = \frac{ - (- 3)}{2} } \\ \\ \longmapsto\sf{ \alpha + \beta = \frac{3}{2} } \: \: \: \: \: \: \: \: \: \:$

Now we find product of the zeros

$\longmapsto\sf{ \alpha \beta = \frac{c}{a} = \frac{5}{2} }$

Using formula to solve :

$\bigstar\bf{ \: \: (x {}^{2} + y {}^{2}) = (x + y) {}^{2} - 2xy }$

By formula putting values

$\longmapsto\sf{ \frac{[( \alpha + \beta ) {}^{2} - 2 \alpha \beta ]}{ \alpha \beta } } \\ \\ \longmapsto\sf{ \frac{[( \frac{3}{2} ) {}^{2} - 2( \frac{5}{2} ) ]}{ \frac{5}{2} } } \\ \\ \longmapsto\sf{ \frac{ \frac{9}{4} - 5 }{ \frac{5}{2} } } \\ \\ \longmapsto\sf{ \frac{ \frac{9 - 20}{4} }{ \frac{5}{2} } } \\ \\ \longmapsto\sf{ \frac{ - 11}{4} \times \frac{2}{5} = \frac{ - 22}{20} }$

${\large{\underline{\boxed{\leadsto{\bf{\red{ \: \frac{ \alpha }{ \beta } + \frac{ \beta }{ \alpha } = \frac{ - 11}{10} }}}}}}}$

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