Math, asked by komalrajputparmar, 7 months ago

7.If the zeroes of the quadratic
polynomial are 2 and -3, then
quadratic equation is
O (A) x2+1x-6=0
O (B) x2+1x+6=0
O (C) x2-1X-6=0
O (D) x2-1x-6=0​

Answers

Answered by Lalitasarate
3

Answer:

-x^2-14x-48=0

Simplifying

-1x2 + -14x + -48 = 0

Reorder the terms:

-48 + -14x + -1x2 = 0

Solving

-48 + -14x + -1x2 = 0

Solving for variable 'x'.

Factor out the Greatest Common Factor (GCF), '-1'.

-1(48 + 14x + x2) = 0

Factor a trinomial.

-1((8 + x)(6 + x)) = 0

Ignore the factor -1.

Subproblem 1

Set the factor '(8 + x)' equal to zero and attempt to solve:

Simplifying

8 + x = 0

Solving

8 + x = 0

Move all terms containing x to the left, all other terms to the right.

Add '-8' to each side of the equation.

8 + -8 + x = 0 + -8

Combine like terms: 8 + -8 = 0

0 + x = 0 + -8

x = 0 + -8

Combine like terms: 0 + -8 = -8

x = -8

Simplifying

x = -8

Subproblem 2

Set the factor '(6 + x)' equal to zero and attempt to solve:

Simplifying

6 + x = 0

Solving

6 + x = 0

Move all terms containing x to the left, all other terms to the right.

Add '-6' to each side of the equation.

6 + -6 + x = 0 + -6

Combine like terms: 6 + -6 = 0

0 + x = 0 + -6

x = 0 + -6

Combine like terms: 0 + -6 = -6

x = -6

Simplifying

x = -6

Solution

x = {-8, -6}

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