Question

Find a quadratic polynomial whose zeros are 6 + root 3 upon 3 and 6 minus root 3 upon 3

Answer 1

Answer:

-1/2 = -1/2

Step-by-step explanation:

To Find -

A quadratic polynomial

Now,

As we know that :-

α + β = -b/a

→ √3/4 + (-2/√3) = -b/a

→ √3/4 - 2/√3 = -b/a

→ 3 - 8/4√3 = -b/a

→ -5/4√3 = -b/a ....... (i)

And

αβ = c/a

→ √3/4 × -2/√3 = c/a

→ -2√3/4√3 = c/a ...... (ii)

Now, From (i) and (ii), we get :

a = 4√3

b = 5

c = -2√3

As we know that :-

For a quadratic polynomial :-

ax² + bx + c

→ (4√3)x² + (5)x + (-2√3)

→ 4√3x² + 5x - 2√3

Verification :-

→ 4√3x² + 5x - 2√3

here,

a = 4√3

b = 5

c = -2√3

Sum of zeroes :-

α + β = -b/a

→ -2/√3 + √3/4 = -(5)/4√3

→ -8 + 3/4√3 = -5/4√3

→ -5/4√3 = -5/4√3

LHS = RHS

And

αβ = c/a

→ -2/√3 × √3/4 = -2√3/4√3

→ -1/2 = -1/2

LHS = RHS

Hence,

Verified...

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Answer 2

$\red{\tt{\underline{\underline{Answer:}}}}$

$\sf{The \ required \ polynomial \ is}$

$\sf{x^{2}-4x+\frac{11}{3}.}$

$\orange{\tt{\underline{\underline{Given:}}}}$

$\sf{\implies{Zeroes \ of \ the \ polynomial}}$

$\sf{are \ \frac{6+\sqrt3}{3} \ and \ \frac{6-\sqrt3}{3}}$

$\pink{\tt{\underline{\underline{To \ find:}}}}$

$\sf{Quadratic \ polynomial}$

$\green{\tt{\underline{\underline{Solution:}}}}$

$\sf{Let \ \alpha \ be \ \frac{6+\sqrt3}{3} \ and}$

$\sf{\beta \ be \ \frac{6-\sqrt3}{3}.}$

$\sf{Sum \ of \ zeroes=\frac{6+\sqrt3}{3}+\frac{6-\sqrt3}{3}}$

$\sf{\therefore{\alpha+\beta=\frac{12}{3}}}$

$\sf{\therefore{\alpha+\beta=4...(1)}}$

$\sf{Product \ of \ zeroes=\frac{6+\sqrt3}{3}\times\frac{6-\sqrt3}{3}}$

$\sf{\therefore{\alpha\beta=\frac{6^{2}-\sqrt3^{2}}{9}}}$

$\sf{\therefore{\alpha\beta=\frac{33}{9}}}$

$\sf{\therefore{\alpha\beta=\frac{11}{3}...(2)}}$

$\sf{Quadratic \ polynomial \ is}$

$\sf{x^{2}-(\alpha+\beta)x+(\alpha\beta)}$

$\sf{\implies{x^{2}-4x+\frac{11}{3}}}$

$\sf\purple{\tt{\therefore{The \ required \ polynomial \ is}}}$

$\sf\purple{\tt{x^{2}-4x+\frac{11}{3}.}}$