Question

Find the roots of the quadratic equation: 2x2 + 3x - 9 = 0?
a. 3/2, -3
b. 2/3, -3
c. 3/2, 3
d. -3/2, -3
e. 3, -3/2

Answer 1

Answer:

d. - 3/2 , - 3

Step-by-step explanation:

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Answer 2

Answer:

Required roots of given equation are $( - 3) \: or \: \frac{3}{2}$

So, option a is the correct option.

Step-by-step explanation:

Given,

$2 {x}^{2} + 3x - 9 = 0$

By middle term method,

$2 {x}^{2} + (6 - 3)x - 9 = 0 \\ 2 {x}^{2} + 6x - 3x - 9 = 0 \\ 2x(x + 3) - 3(x + 3) = 0 \\ (x + 3)(2x - 3) = 0$

We know,if product of two terms is zero then they are separately zero.

So,

$x + 3 = 0 \\ x = - 3$

and

$2x - 3 = 0 \\ 2x = 3 \\ x = \frac{3}{2}$

So, required roots of given equation are

$( - 3) \: or \: \frac{3}{2}$

This is a problem of Algebra.

Some important Algebra formulas:

${(x + y)}^{2} = {x}^{2} + 2xy + {y}^{2} \\ {(x - y)}^{2} = {x}^{2} - 2xy + {y}^{2} \\ {(x + y)}^{3} = {x}^{3} + 3 {x}^{2} y + 3x {y}^{2} + {y}^{3} \\ {(x - y)}^{3} = {x}^{3} - 3 {x}^{2} y + 3x {y}^{2} - {y}^{3} \\ {x}^{3} + {y}^{3} = {(x + y)}^{3} - 3xy(x + y) \\ {x}^{3} - {y}^{3} = {(x - y)}^{3} + 3xy(x - y) \\ {x}^{2} - {y}^{2} = (x + y)(x - y) \\ {x}^{2} + {y}^{2} = {(x - y)}^{2} + 2xy \\ {x}^{2} - {y}^{2} = {(x + y)}^{2} - 2xy \\ {x}^{3} - {y}^{3} = (x - y)( {x}^{2} + xy + {y}^{2} ) \\ {x}^{3} + {y}^{3} = (x - + y)( {x}^{2} - xy + {y}^{2} )$

Know more about Algebra,

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2) https://brainly.in/question/1169549

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