Question

Question-


If alpha and beta are the zeros of quadratic p(x) such alpha+beta=4 and alpha-beta-8.find a quadratic p(x) having alpha and beta as its zero.


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Answer 1

Explanation:

Given,

$\alpha + \beta = 4$

$\alpha - \beta = 8$

In this question, we can simply find the roots and then, form the equation.

$( \alpha + \beta ) + ( \alpha - \beta ) = 4 + 8$

$2 \alpha = 12$

$\alpha = \frac{12}{2} = 6$

Therefore,

$\beta = 4 - 6 = - 2$

In a quadratic polynomial of form ax^2 + bx + c with alpha and beta as roots,

$\alpha + \beta = \frac{ - b}{a}$

Hence, in this case,

$\alpha + \beta = 4 \: \: (given)$

Assuming a is equal to 1,

$- b = 4$

$b = - 4$

It is also known that

$\alpha \beta = \frac{c}{a}$

$c = (6)( - 2) = - 12$

Hence, the quadratic polynomial to be formed is

$x {}^{2} - 4x - 12$

This is when a is considered to be 1. If a is to be taken as any other number, the following quadratic equation will be formed:

$a {x}^{2} - (a)4x - (a)12$

Answer: x^2 - 4x - 12

I hope this answers helps! :D

Answer 2

$\huge\bold{\textbf{\textsf{{\color{cyan}{Answer}}}}}∴$

Hello people >•<

Given,

$\alpha + \beta = 4α+β=4</p><p>\alpha - \beta = 8α−β=8$

In this question, we can simply find the roots and then, form the equation.

$( \alpha + \beta ) + ( \alpha - \beta ) = 4 + 8(α+β)+(α−β)=4+8</p><p>2 \alpha = 122α=12$

$\alpha = \frac{12}{2} = 6α=212=6$

Therefore,

$\beta = 4 - 6 = - 2β=4−6=−2 \: </p><p>In \: a \: quadratic \: polynomial \: of \: form \: ax^2 + bx + c \: with \: alpha \: and \: beta \: as \: roots,$

$\alpha + \beta = \frac{ - b}{a}α+β=a−b$

Hence, in this case,

$\alpha + \beta = 4 \: \: (given)α+β=4(given)</p><p>$

Assuming a is equal to 1,

$- b = 4−b=4 \\ </p><p>b = - 4b=−4$

It is also known that

$\alpha \beta = \frac{c}{a}αβ=ac</p><p>c = (6)( - 2) = - 12c=(6)(−2)=−12$

Hence, the quadratic polynomial to be formed is

$x {}^{2} - 4x - 12x2−4x−12$

This is when a is considered to be 1. If a is to be taken as any other number, the following quadratic equation will be formed:

$a {x}^{2} - (a)4x - (a)12ax2−(a)4x−(a)12$