Question

pls solve this 100 point question and I will make you brainLOST

eq is x-square given in question​

Answer 1

Given:

• A quadratic equation x²+px+12=0 whose roots are a and b
• a-b=1

To Find:

• Value of 'p'

Solution:

We know that,

$\rightarrow\sf{Sum\:of\:Zeroes=\dfrac{-Coefficient\:of\:x}{Coefficient\:of\:x^{2}}}$

$\rightarrow\sf{Product\:of\:Zeroes=\dfrac{Constant\:Term}{Coefficient\:of\:x^{2}}}$

For given quadratic equation,

$\longrightarrow\sf{Product\:of\:Zeroes=\dfrac{12}{1}}$

$\longrightarrow\sf{ab=12}$

$\longrightarrow\sf{b=\dfrac{12}{a}}---------(1)$

$\longrightarrow\sf{Sum\:of\:Zeroes=\dfrac{-p}{1}}$

$\longrightarrow\sf{a+b=-p}$

From (1)

$\longrightarrow\sf{a+\dfrac{12}{a}=p}-------(2)$

Now, it is given that

$\longrightarrow\sf{a-b=1}$

From (1)

$\longrightarrow\sf{a-\dfrac{12}{a}=1}$

$\longrightarrow\sf{\dfrac{a^{2}-12}{a}=1}$

$\longrightarrow\sf{a^{2}-12=a}$

$\longrightarrow\sf{a^{2}-a-12=0}$

$\longrightarrow\sf{a^{2}+3a-4a-12=0}$

$\longrightarrow\sf{a(a+3)-4(a+3)=0}$

$\longrightarrow\sf{(a-4)(a+3)=0}$

$\longrightarrow\sf{a=4,-3}$

On putting values of a in (2), we get

Case 1: When a=4

$\longrightarrow\sf{-p=4+\dfrac{12}{4}}$

$\longrightarrow\sf{-p=4+3}$

$\longrightarrow\sf\pink{-p=7}$

$\longrightarrow\sf\pink{p=-7}$

Case 2: When a= -3

$\longrightarrow\sf{-p=-3+\dfrac{12}{-3}}$

$\longrightarrow\sf{-p=-3-4}$

$\longrightarrow\sf\purple{-p=-7}$

$\longrightarrow\sf\pink{p=7}$

Hence, the correct answer is option (D) i.e. value of p is 7 or -7.

Answer 2

$\red{ANSWER}$

Sum of roots = a + b = -p.

Product of roots = ab = 12.

(a - b)^2 = (a + b)^2 - 4(ab) = p^2 - 48 = 1. ==>*** p = ±7.***

Case 1: x^2 + 7x + 12 = (x+4)(x+3)=0. x = -4 or -3.

Case 2: x^2 - 7x + 12 = (x -4)(x-3)=0. x = 4 or 3