Question

2. Find the zeros of the polynomial x² + x - p(p+1).​

Answer 1

Given

• f(x)=x² + x - p(p+1)

To Find

• Zeros of the polynomial

Solution

______Meathod 1 ______

• x²+x-p(p+1)=0

• On factorising we get,

• x²+(p+1)x-px-p(p+1)=0

• x(x+p+1) -p(x+p+1)=0

• (x+p+1)(x-p)=0

• x=-p-1, and x= p

The zeros of polynomial are: p-1 and p.

______Meathod 2 ______

$\sf \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: x²+x-p(p+1)=0$

On Comparing with ax²+bx+c = 0, we get,

$\: \: \: \: \\ \sf a= 1 \: \: , \: \: b = 1 \: , \: c = -p(p+1)$

By Quadratic Equation

$\sf~~~\implies~~~~ x = \dfrac{-b±\sqrt{b²-4ac}}{2a}$

• Putting the values

$\sf~~~\implies~~~~ x = \dfrac{-1±\sqrt{1²-4 \times 1 \times - p(p + 1)}}{2 \times 1}$

$\sf~~~\implies~~~~ x = \dfrac{-1±\sqrt{1-4 \times \{ - {p}^{2} - p \}}}{2 }$

$\sf~~~\implies~~~~ x = \dfrac{-1±\sqrt{1 + 4 {p}^{2} + 4p }}{2 }$

$\sf~~~\implies~~~~ x = \dfrac{-1±\sqrt{ 4 {p}^{2} + 4p + 1}}{2 }$

$\sf~~~\implies~~~~ x = \dfrac{-1±\sqrt{ (2 {p})^{2} + 2(2p)1 + {1}^{2} }}{2 }$

• Using identity a²+2ab+b² = (a+b)²

$\sf~~~\implies~~~~ x = \dfrac{-1±\sqrt{ { (2p + 1) }^{2} }}{2 }$

$\sf~~~\implies~~~~ x = \dfrac{-1±{ (2p + 1) }}{2 }$

$\sf~~~~~~~ x = \dfrac{-1 + { (2p + 1) }}{2 } \: \: \: \: \: \: \: \bigg| \: \: \: \: \: x = \dfrac{-1 - { (2p + 1) }}{2 }$

$\sf~~~~~~~ x = \dfrac{-1 + 2p + 1}{2 } \: \: \: \: \: \: \: \bigg| \: \: \: \: \: x = \dfrac{-1 + 2p - 1}{2 }$

$\sf~~~~~~~ x = \dfrac{ 2p }{2 } \: \: \: \: \: \: \: \bigg| \: \: \: \: \: x = \dfrac{ 2p - 2}{2 }$

$\sf~~~~~~~ x = p \: \: \: \: \: \: \: \bigg| \: \: \: \: \: x = \dfrac{ 2(p - 1)}{2 }$

$\sf~~~~~~~ x = p \: \: \: \: \: \: \: \bigg| \: \: \: \: \: x = p - 1$

The zeros of polynomial are: p and p-1 .

Answer 2

$\huge \mathbb{ \red {★᭄ꦿ᭄S} \pink{ᴏ}\purple{ʟᴜ} \blue {ᴛ} \orange{ɪ} \green{ᴏɴ★᭄ꦿ᭄}}$

$\tt \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: x²+x-p(p+1)=0$

• ↪On Comparing with ax²+bx+c = 0, we get

$\begin{gathered}\sf a= 1 \: \: , \: b = 1 \: , \: c = -p(p+1) \end{gathered}$

BY QUADRATIC EQYATION

$\tt~~~\implies~~~~ x = \dfrac \red{-b±\sqrt{b²-4ac}} \pink{2a}$

• ↪BY PUTTING THE VALUES

$\tt~~~\implies~~~~ x = \dfrac \blue{-1±\sqrt{1²-4 \times 1 \times - p(p + 1)}} \purple{2 \times 1}$

$\tt~~~\implies~~~~ x = \dfrac \red{-1±\sqrt{1-4 \times \{ - {p}^{2} - p \}}} \orange{2 }$

$\tt~~~\implies~~~~ x = \dfrac \pink{-1±\sqrt{1 + 4 {p}^{2} + 4p }} \purple{2 }$

$\tt~~~\implies~~~~ x = \dfrac \blue{-1±\sqrt{ 4 {p}^{2} + 4p + 1}} \green{2 }$

$\tt~~~\implies~~~~ x = \dfrac \purple{-1±\sqrt{ (2 {p})^{2} + 2(2p)1 + {1}^{2} }} \red{2 }$

• ↪Using identity a²+2ab+b² = (a+b)²

$\tt~~~\implies~~~~ x = \dfrac \orange{-1±\sqrt{ { (2p + 1) }^{2} }} \blue{2 }$

$\tt~~~\implies~~~~ x = \dfrac \red{-1±{ (2p + 1) }} \green{2 }$

$\tt~~~~~~~ x = \dfrac \purple{-1 + { (2p + 1) }} \red{2 } \: \: \: \: \: \: \: \bigg| \: \: \: \: \: x = \dfrac \green{-1 - { (2p + 1) }} \pink{2 }$

$\tt~~~~~~~ x = \dfrac \green{-1 + 2p + 1} \red{2 } \: \: \: \: \: \: \: \bigg| \: \: \: \: \: x = \dfrac \orange{-1 + 2p - 1} \purple{2 }$

$\tt~~~~~~~ x = \dfrac \green{ 2p } \pink{2 } \: \: \: \: \: \: \: \bigg| \: \: \: \: \: x = \dfrac \red{ 2p - 2} \orange{2 }$

$\sf~~~~~~~ x = \blue{p} \: \: \: \: \: \: \: \bigg| \: \: \: \: \: x = \dfrac \red{ 2(p - 1)} \pink{2}$

$\begin{gathered}\tt~~~~~~~ x = \pink{p} \: \: \: \: \: \: \: \bigg| \: \: \: \: \: x = \red{p - 1} \\ \\ \end{gathered}$

• The zeros of polynomial are: p and p-1 .✅