Question

Find a quadratic polynomial, the sum and product of whose zeroes are √2 and -3/2, respectively. Also find its zeroes and verify the relationship between the zeroes and the coefficients.​

Answer 1

Given:

• The sum of quadratic polynomial is √2
• The Product of quadratic polynomial is -3/2

Need to Find:

• Quadratic polynomial
• Zeroes of Quadratic polynomial

To Verify:

• Relationship between zeroes and the Coefficient.

Explanation:

Here, we are given the sum and product of Quadratic polynomial. We have to find the Quadratic polynomial. Also, We have ti find it's zeroes. For finding the quadratic polynomial. We have to take α and β as zeroes of polynomial. We know, if α and β are the zeroes of a quadratic polynomial. Then, polynomial f(x) is given by f(x)=x²-(α+β)x+αβ. Equating all the values we get, the required polynomial. We also know, zeroes of f(x) is given by f(x)=0. After futher solving we will get the zeroes of polynomial. We also have to verify the relationship between the zeroes and coefficient of the polynomial. For verifying the relationship. We know, α+β=-(Coefficient of x)/(Coefficient of x²) and αβ=(Constant term)/(Coefficient of x²)

__________________________

Solution:

Let, α and β be the zeroes of required polynomial. It is given that sum of zeroes=√2 and Product of zeroes=-3/2. So,

$\leadsto\sf{\alpha+\beta=\sqrt{2}}$

$\leadsto\sf{\alpha\times \beta=\large{\frac{-3}{2}}}$

We know,

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

The quadratic polynomial is::

$\rightarrow\sf{f(x)={x}^{2}-\sqrt{2}x-\large{\frac{-3}{2}}}$

Now,

$\rightarrow\sf{f(x)={x}^{2}-\sqrt{2}x-\large{\frac{-3}{2}}}$

$\rightarrow\sf{f(x)=\large{\frac{1}{2}}({2x}^{2}-2\sqrt{2}x-3)}$

$\rightarrow\sf{f(x)=\large{\frac{1}{2}}({2x}^{2}-3\sqrt{2}x+\sqrt{2}x-3)}$ $\\\rightarrow\sf{f(x)=\large{\frac{1}{2}}[\sqrt{2}x(\sqrt{2}x-3)+(\sqrt{2}x-3)]}$

$\rightarrow\bold{f(x)=\large{\frac{1}{2}}(\sqrt{2}x-3)(\sqrt{2}x+1)}$

The zeroes of f(x) are given by f(x)=0. So,

$\dashrightarrow\sf{f(x)=0}$

$\dashrightarrow\sf{\large{\frac{1}{2}}(\sqrt{2}x-3)(\sqrt{2}x+1)=0}$

$\dashrightarrow\sf{\sqrt{2}x-3=0\: \: or\: \: \sqrt{2}x+1=0}$

$\dashrightarrow\sf{x=\large{\frac{3}{\sqrt{2}}}\: or\: x=\large{\frac{-1}{\sqrt{2}}}}$

Hence, the zeroes of f(x) are 3/√2 and -1/√2

Verification:

We are given,

$\implies\sf{Sum\: of\: zeroes=\sqrt{2}}$

Now,

$\implies\sf{-\large{\frac{Coefficient\: of\: x}{Coefficient\: of\: {x}^{2}}}=-\large{\frac{\sqrt{2}}{1}}}$

$\implies\sf{-\large{\frac{Coefficient\: of\: x}{Coefficient\: of\: {x}^{2}}}=\sqrt{2}}$

Hence, $\\ \implies\bold{-\frac{Coefficient\: of\: x}{Coefficient\: of\: {x}^{2}}=Sum\: of\: zeroes}$

Also, we are given with,

$\implies\sf{Product\: of\: zeroes=-\large{\frac{\sqrt{3}}{2}}}$

Now,

$\implies\sf{-\large{\frac{Constant\: term}{Coefficient\: of\: {x}^{2}}}=-\large{\frac{{}^{3}\! /{}_{2}}{1}}}$

$\implies\sf{-\large{\frac{Constant\: term}{Coefficient\: of\: {x}^{2}}}=-\large{\frac{3}{2}}}$

Hence, $\\\implies\bold{-\frac{Constant\: term}{Coefficient\: of\: {x}^{2}}=Product\: of\: zeroes}$

Hence, Verified!

Answer 2

Answer:

Given:

The sum of quadratic polynomial is √2

The Product of quadratic polynomial is -3/2

Need to Find:

Quadratic polynomial

Zeroes of Quadratic polynomial

To Verify:

Relationship between zeroes and the Coefficient.

Explanation:

Here, we are given the sum and product of Quadratic polynomial. We have to find the Quadratic polynomial. Also, We have ti find it's zeroes. For finding the quadratic polynomial. We have to take α and β as zeroes of polynomial. We know, if α and β are the zeroes of a quadratic polynomial. Then, polynomial f(x) is given by f(x)=x²-(α+β)x+αβ. Equating all the values we get, the required polynomial. We also know, zeroes of f(x) is given by f(x)=0. After futher solving we will get the zeroes of polynomial. We also have to verify the relationship between the zeroes and coefficient of the polynomial. For verifying the relationship. We know, α+β=-(Coefficient of x)/(Coefficient of x²) and αβ=(Constant term)/(Coefficient of x²)

__________________________

Solution:

Let, α and β be the zeroes of required polynomial. It is given that sum of zeroes=√2 and Product of zeroes=-3/2. So,

\leadsto\sf{\alpha+\beta=\sqrt{2}}⇝α+β=2

\leadsto\sf{\alpha\times \beta=\large{\frac{-3}{2}}}⇝α×β=2−3

We know,

\rightarrow\sf{f(x)={x}^{2}-(\alpha+\beta)x+\alpha\beta}→f(x)=x2−(α+β)x+αβ

The quadratic polynomial is::

\rightarrow\sf{f(x)={x}^{2}-\sqrt{2}x-\large{\frac{-3}{2}}}→f(x)=x2−2x−2−3

Now,

\rightarrow\sf{f(x)={x}^{2}-\sqrt{2}x-\large{\frac{-3}{2}}}→f(x)=x2−2x−2−3

\rightarrow\sf{f(x)=\large{\frac{1}{2}}({2x}^{2}-2\sqrt{2}x-3)}→f(x)=21(2x2−22x−3)

\rightarrow\sf{f(x)=\large{\frac{1}{2}}({2x}^{2}-3\sqrt{2}x+\sqrt{2}x-3)}→f(x)=21(2x2−32x+2x−3) \begin{lgathered}\\\rightarrow\sf{f(x)=\large{\frac{1}{2}}[\sqrt{2}x(\sqrt{2}x-3)+(\sqrt{2}x-3)]}\end{lgathered}→f(x)=21[2x(2x−3)+(2x−3)]

\rightarrow\bold{f(x)=\large{\frac{1}{2}}(\sqrt{2}x-3)(\sqrt{2}x+1)}→f(x)=21(2x−3)(2x+1)

The zeroes of f(x) are given by f(x)=0. So,

\dashrightarrow\sf{f(x)=0}⇢f(x)=0

\dashrightarrow\sf{\large{\frac{1}{2}}(\sqrt{2}x-3)(\sqrt{2}x+1)=0}⇢21(2x−3)(2x+1)=0

\dashrightarrow\sf{\sqrt{2}x-3=0\: \: or\: \: \sqrt{2}x+1=0}⇢2x−3=0or2x+1=0

\dashrightarrow\sf{x=\large{\frac{3}{\sqrt{2}}}\: or\: x=\large{\frac{-1}{\sqrt{2}}}}⇢x=23orx=2−1

Hence, the zeroes of f(x) are 3/√2 and -1/√2

Verification:

We are given,

\implies\sf{Sum\: of\: zeroes=\sqrt{2}}⟹Sumofzeroes=2

Now,

\implies\sf{-\large{\frac{Coefficient\: of\: x}{Coefficient\: of\: {x}^{2}}}=-\large{\frac{\sqrt{2}}{1}}}⟹−Coefficientofx2

Step-by-step explanation:

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