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For someone who is not comfortable with Quadratic Equations and forming factors, an easy way out would be to use the answer options to get the answer. :)

2x^2 + x -2 = 1

A. \(2(-6)^2 + (-6) - 2\) is not equal to \(1\)
B. \(2(-1)^2 + (-1) - 2\) is not equal to \(1\)
C. \(2(1)^2 + (1) - 2\) is equal to \(1\)
D. \(2(3/2)^2 + (3/2) - 2\) is not equal to \(1\)
E. \(2(3)^2 + (3) - 2\) is not equal to \(1\)

Hence Option C is the correct answer.

Note : Once you confirm that Option C gives us 1, we do not need to check Option D and E. :)

Thanks,
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2x2+x−2−1=02x2+x−2−1=0
2x2+x−3=02x2+x−3=0
2x2+3x−2x−3=02x2+3x−2x−3=0
x(2x+3)−1(2x+3)=0x(2x+3)−1(2x+3)=0
(x−1)(2x+3)=0(x−1)(2x+3)=0
x−1=0x−1=0
x=1

Ans:-C
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Bunuel
Which of the following is a solution for 2x^2 + x − 2 = 1?

A. −6
B. −1
C. 1
D. 3/2
E. 3

Let’s simplify the given equation:

2x^2 + x − 2 = 1

2x^2 + x − 3 = 0

(2x + 3)(x − 1) = 0

2x + 3 = 0

2x = -3

x = -3/2

OR

x - 1 = 0

x = 1

Answer: C
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Bunuel
Which of the following is a solution for 2x^2 + x − 2 = 1?

A. −6
B. −1
C. 1
D. 3/2
E. 3

\(2x^2 + x − 2 = 1\)

\(2x^2 + x − 2 − 1 = 0\)

\(2x^2 + x − 3 = 0\)

now all i have to do now is test values

clearly only 1 yields 0

\(2 *1^2 + 1 − 3 = 0\)

\(3-3=0\)
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Bunuel
Which of the following is a solution for 2x^2 + x − 2 = 1?

A. −6
B. −1
C. 1
D. 3/2
E. 3


Hey,

PFB the solution.

    \(2x^2 + x - 2 = 1\)

The above equation can be written as -

    \(2x^2 + x -2 -1 = 0\)
    \(2x^2 + x - 3 = 0\)
    \(2x^2 + 3x - 2x - 3 = 0\)
    \(x( 2x + 3) -1 (2x +3) = 0\)
    \((x - 1) (2x + 3) = 0\)

Therefore either \(x-1 = 0\) which gives \(x =1\) or
\(2x +3 = 0\), which gives \(x = -\frac{3}{2}\)

Since the question asks us, which of the following could be a solution of \(x\), we can confidently say that the answer will be 1, Option C

Thanks,
Saquib
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Hello!

Does anyone know which is the basic theory to know how to solve this kind of QE?

    \(2x^2 + x -2 -1 = 0\)
    \(2x^2 + x - 3 = 0\)
    \(2x^2 + 3x - 2x - 3 = 0\)
    \(x( 2x + 3) -1 (2x +3) = 0\)
    \((x - 1) (2x + 3) = 0\)

Kind regards!
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