Question

find a quadratic polynomial where zeroes are -2upon root 3 and root 3 upon 4​

Answer 1

Step-by-step explanation:

Given -

• Zeroes of the polynomial are -2/√3 and √3/4

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...

Answer 2

$\large{\underline{\bf{\green{Given:-}}}}$

✰ Zeroes of required polynomial is given as -2/√3 and √3/4

$\large{\underline{\bf{\green{To\:Find:-}}}}$

✰ we need to find the quadratic polynomial.

$\huge{\underline{\bf{\red{Solution:-}}}}$

Let the quadratic polynomial be

⠀⠀⠀⠀⠀ax² + bx + c.

and it's zeroes are α and β Then,

α = -2/√3

β = √3/4

So,

Sum of zeroes :-

(α + β) = $\sf\:\frac{-2}{\sqrt{3}}+\frac{\sqrt{3}}{4}$

$: \implies \sf\:\frac{-5}{4\sqrt{3}}$

product of zeroes:-

αβ = $\sf\:\frac{-2}{\sqrt{3}}\times\frac{\sqrt{3}}{4}$

$: \implies \sf\:\frac{-1}{2}$

So,

Quadratic polynomial:-

$: \implies \sf\:x^2-(\alpha+\beta)x+\alpha\beta$

$: \implies \sf\: \frac{4 \sqrt{3} {x}^{2} + 5x - 2 \sqrt{3} }{4 \sqrt{3} } = 0 \\ \\ : \implies \sf\: 4 \sqrt{3} {x}^{2} + 5x - 2 \sqrt{3} = 0$

Hence,

the quadratic polynomial is

$\bf\: 4 \sqrt{3} {x}^{2} + 5x - 2 \sqrt{3} = 0$

Verification:-

Sum of zeroes :- -b/a

$: \implies \sf\:\frac{-2}{\sqrt{3}}+\frac{\sqrt{3}}{4}=\frac{-5}{4\sqrt{3}}$

$: \implies \sf\:\frac{-8+3}{4\sqrt{3}}=\frac{-5}{4\sqrt{3}}$

$: \implies \sf\:\frac{-5}{4\sqrt{3}}=\frac{-5}{4\sqrt{3}}$

Product of zeroes :- c/a

$: \implies \sf\:\frac{-2}{\sqrt{3}}\times\frac{\sqrt{3}}{4}=\frac{-2\sqrt{3}}{4\sqrt{3}}$

$: \implies \sf\:\frac{-1}{2}=\frac{-1}{2}$

LHS = RHS

Hence verified

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