Question

Find the quadratic polynomial in each case (i) sum and product of the zeroes respectively
are _1/4 and 1/4
and (ii) for the zeroes a, ß are 2, - 1.​

Answer 1

▪IN THE FIRST QUESTION :

☯GIVEN :

• $\alpha + \beta = \dfrac{-1}{4}$

• $\alpha \beta = \dfrac{1}{4}$

☯TO FIND :

• Quadratic Polynomial.

☯FORMULA REQUIRED :

• $\underline{\boxed{\purple{\bf{x^2- \left(\alpha + \beta\right)x + \alpha\beta =0 }}}}$

➲SOLUTION :

• By substituting the values, $\bf{\alpha + \beta =\dfrac{-1}{4}}$ and $\bf{\alpha \beta = \dfrac{1}{4}}$

$:\implies {\sf x^2 - \left(\alpha + \beta \right)x + \alpha \beta = 0}$

$:\implies {\sf x^2- (\dfrac{-1}{4} )x+\dfrac{1}{4} =0 }$

$:\implies {\sf x^2+ \dfrac{1}{4}x+\dfrac{1}{4} =0 }$

$:\implies {\sf \dfrac{4x^2 +x + 1}{4} =0 }$

• By cross multiplication :

$:\implies {\sf 4x^2+x + 1 = 0 \times 4}$

$:\implies{ \underline{ \boxed{ \blue{\bf{ 4 x^2 +x + 1 = 0}}}}}$

$\huge{\green{\therefore}}$Quadratic Polynomial = $\bf{ 4x^2 + x + 1 }$

▪IN THE SECOND QUESTION :

☯GIVEN :

• $\bf{\alpha = 2}$

• $\bf{\beta = -1 }$

☯TO FIND :

• Quadratic Polynomial.

☯FORMULA REQUIRED :

• $\underline{\boxed{\purple{\bf{x^2- \left(\alpha + \beta\right)x + \alpha\beta =0 }}}}$

➲SOLUTION :

★ Sum of Zeros :

$: \leadsto \alpha + \beta$

$: \leadsto2 + (-1)$

$: \leadsto 2-1$

$: \leadsto \bf{1}$

★ Product of Zeros :

$: \leadsto \alpha \beta$

$: \leadsto 2 \times (-1)$

$: \leadsto \bf{ -2 }$

• By substituting the values, $\bf{\alpha + \beta = 1}$ and $\bf{\alpha \beta = -2}$

$:\implies {\sf x^2 - \left(\alpha + \beta \right)x + \alpha \beta = 0}$

$:\implies {\sf x^2- \left( 1\right )x+(-2) =0 }$

$: \implies \underline{\boxed{ \purple{\bf{ x ^2 - x - 2 = 0 }}}}$

$\huge{\green{\therefore}}$ Quadratic polynomial = $\bf{ x ^2 - x -2}$