Math, asked by yashisharma14402, 5 hours ago

Find a quadratic polynomial, the sum and product of whose zeroes are √3 and1/√3respectively

Answers

Answered by amansharma264
12

EXPLANATION.

Quadratic polynomial,

Sum of the zeroes = √3.

Products of its zeroes = 1/√3.

As we know that,

Sum of the zeroes of the quadratic polynomial.

⇒ α + β = - b/a.

⇒ α + β = √3.

Products of the zeroes of the quadratic polynomial.

⇒ αβ = c/a.

⇒ αβ = 1/√3.

As we know that,

Formula of quadratic polynomial.

⇒ x² - (α + β)x + αβ.

Put the values in the equation, we get.

⇒ x² - (√3)x + (1/√3) = 0.

⇒ √3x² - 3x + 1 = 0.

                                                                                                                       

MORE INFORMATION.

Nature of th roots of the quadratic expression.

(1) = Real and unequal, if b² - 4ac > 0.

(2) = Rational and different, if b² - 4ac is a perfect square.

(3) = Real and equal, if b² - 4ac = 0.

(4) = If D < 0 Roots are imaginary and unequal Or complex conjugate.

Answered by BrainlyArnab
0

 \huge{ \sqrt{3}  {x}^{2}  - 3x + 1}

Step-by-step explanation:

Q.

Find a quadratic polynomial, the sum and product of whose zeroes are 3 and 1/3 respectively

.

Solution -

sum \: of \: zeroes =   - \frac{b}{a}  \\  =  &gt;  \alpha  +  \beta  =  \sqrt{3}  \\  \\ product \: of \: zeroes =  \frac{c}{a}  \\  =  &gt;  \alpha  \beta  =  \frac{1}{ \sqrt{3} }

Using the quadratic equation formula

 {x}^{2}   - ( \alpha  +  \beta )x + ( \alpha  \beta )  = 0 \\  \\  =  &gt;  {x}^{2}  -  \sqrt{3} x +  \frac{1}{ \sqrt{3} }  = 0 \\  \\ (multiply \: by \sqrt{3} ) \\  =  &gt;  \sqrt{3} ( {x}^{2}  -  \sqrt{3} x +  \frac{1}{\sqrt{3} }  = 0) \\  \\  =  &gt;    \underline \red{\sqrt{3}  {x}^{2}  - 3x + 1 = 0}

Hence the polynomial is 3x² - 3x + 1 .

hope it helps.

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