Question

if alfa and beta are two zeroes of the quadratic polynomial p(x) =2xsquare-3x+7 find (i) 1/alfa+1/beta (ii) alfa square+ beta square​

Answer 1

EXPLANATION.

If α and β are two zeroes of the quadratic polynomial.

⇒ 2x² - 3x + 7.

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 = 7/2.

To find :

⇒ 1/α + 1/β.

⇒ β + α/αβ.

⇒ 3/2/7/2.

⇒ 3/2 x 2/7 = 3/7.

⇒ α² + β².

⇒ [α + β]² - 2αβ.

⇒ [3/2]² - 2[7/2].

⇒ [9/4] - 7.

⇒ 9 - 28/4.

⇒ -19/4.

Values of 1/α + 1/β = 3/7.

Values of α² + β² = -19/4.

                                                                                                                           

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

Given:-

• α and β are two zeroes of the quadratic polynomial 2x² - 3x + 7.

To Find:-

• $\dfrac{1}{α}+\dfrac{1}{β}$, and

• α² + β²

Solution:-

On comparing the given equation to the standard equation i.e. ax² + bx + c we get,

• a = 2
• b = -3
• c = 7

Sum of zeroes = α + β = $\dfrac{-b}{a}$

$⟹\:α + β = \dfrac{-(-3)}{2}$

${\boxed{⟹\:α + β = \dfrac{3}{2}}}$

Product of Zeroes = αβ = $\dfrac{c}{a}$

${\boxed{⟹\:αβ = \dfrac{7}{2}}}$

Now, Solving for ${\bold{\dfrac{1}{α}+\dfrac{1}{β}}}$

$=\:\dfrac{1}{α}+\dfrac{1}{β}$

$=\:\dfrac{β+α}{αβ}$

Putting the values,

$=\:\dfrac{\dfrac{3}{2}}{\dfrac{7}{2}}$

$=\:\dfrac{3}{7}$ Ans.

Now, Solving for α² + β²

Using Identity, (a + b)² = a² + b² + 2ab

$⟹\:(α+ β)² = α² + β²+ 2αβ$

Putting values,

$⟹\:(\dfrac{3}{2})² = α² + β²+ 2×\dfrac{7}{2}$

$⟹\:\dfrac{9}{4} -7= α² + β²$

$⟹\:α² + β² = \dfrac{9-28}{4}$

$⟹\:α² + β² = \dfrac{-19}{4}$ Ans.

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