Question

Find the quadratic polynomial if the sum of its zeros is minus one upon root 2 and the product of zeros is 2 upon 3

Answer 1

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

• Sum of zeroes = -1/√2
• product of zeroes = 2/3

$\large\bf\underline {To \: find:-}$

• Quadratic polynomial

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

Let α and β are the zeroes of the required polynomial.

• α + β = -1/√2
• αβ = 2/3

Formula for quadratic polynomial:-

$\large \boxed{ \bf {x}^{2} - ( \alpha + \beta )x + \alpha \beta }$

$\dashrightarrow \rm \: {x}^{2} - ( \frac{ - 1}{ \sqrt{2} } )x + \frac{2}{3} \\ \\ \dashrightarrow \rm \: {x}^{2} + \frac{x}{ \sqrt{2} } + \frac{2}{3} \\ \\ \dashrightarrow \rm \: \frac{ 3\sqrt{2} {x}^{2} + 3 x + 2 \sqrt{2} }{ 3\sqrt{2} } = 0 \\ \\ \dashrightarrow \rm \: 3\sqrt{2} {x}^{2} + 3x + 2 \sqrt{2} = 0$

So,

The quadratic polynomial is:-

• 3√2x² + 3x + 2√2

$\large\underline{\bf\:Verification:-}$

p(x) = 3√2x² +3x +2√2

• a = 3√2
• b = 3
• c = 2√2

≫ Sum of zeroes = -b/a

»» -1/√2 = -3/3√2

»» -1/2 = -1/√2

≫ Product of zeroes = c/a

»» 2/3 = 2√2/3√2

»» 2/3 = 2/3

LHS = RHS

Hence Verified.

Answer 2

◘ Given ◘

• Sum of the zeros of a quadratic polynomial is -1/√2. And,
• Product of the zeros of the quadratic polynomial is 2/3.

Let the zeros be α and β. Then,

$\bigstar$ α + β = -1/√2

$\bigstar$ αβ = 2/3

$\rule{170}4$

◘ Objective ◘

To find the quadratic polynomial whose sum of zeros and product of zeros are given.

$\rule{170}4$

◘ Calculation ◘

Let S = α + β = -1/√2, and P = αβ = 2/3.

We know that a quadratic polynomial is of the form :-

$\sf{x^2-(S)x+P}$

On putting the values, we get the required polynomial as :-

$\sf{p(x)=x^2-\bigg(\dfrac{-1}{\sqrt{2}}\bigg)x +\dfrac{2}{3} }\\\\\sf{\to p(x)=x^2+\dfrac{1}{\sqrt{2}}x+\dfrac{2}{3} }\\\\\boxed{\sf{\color{magenta}{\to p(x)=x^2+\dfrac{x}{\sqrt{2}}+\dfrac{2}{3} }}}$