Question

If p = 1 and q = –2 are roots of a quadratic equation, then quadratic equation will be ……………

1️⃣ x² + 2x –1= 0
2️⃣ x² – x – 2 = 0
3️⃣ x² – 2x + 1= 0
4️⃣ x² + x + 2 = 0.​

Answer 1

Answer:

x²+x-2=0

Step-by-step explanation:

general form = x²-x(α+β)+αβ

be 1=α and -2=β

α+β=1+(-2)= -1

αβ = (1)(-2) =-2

x²-(-1)+(-2)=0

x²+1x-2=0

Answer 2

$\huge\bf\pink{Question}$

If the values p = 1 and q = –2 are the roots of the quadratic equation then the quadratic equation is *

1️⃣ x² + 2x –1 = 0

2️⃣ x² - x - 2 = 0

3️⃣ x² - 2x + 1 = 0

4️⃣ x² + x + 2 = 0

$\huge\bf\pink{Answer}$

$\sf\green{Given:-}$Zeros of a polynomial are 1 and - 2

$\sf\green{To\:Find:-}$The Quadratic equation

$\sf\green{Proof:-}$

We know that,

Any Quadratic equation given with roots $\alpha$ and $\beta$ is of the form,

$\boxed{\red{\sf x^2 \: - \: (\alpha \: + \: \beta) \: x \: + \: \alpha \beta \: = \: 0}}$

Here,

With given zeros p and q,

$\implies{\red{\sf x^2 \: - \: (p \: + \: q) \: x \: + \: pq \: = \: 0}}$

Given that,

p = 1

q = - 2

Substituting the values,

x² - (1 + (- 2)) x + 1 * - 2 = 0

x² - (1 - 2) x - 2 = 0

x² - (- 1) x - 2 = 0

x² + 1x - 2 = 0

x² + x - 2 = 0

$\therefore\bf\red{Quadratic \: equation \: = \: x^2 \: + \: x \: - \: 2 \: = \: 0}$