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What are the solutions to this quadratic equation?
[tex]\[ -2x^2 + x + 4 = 0 \][/tex]

A. [tex]\( x = \frac{-1 \pm \sqrt{33}}{4} \)[/tex]
B. [tex]\( x = \frac{1 \pm \sqrt{33}}{4} \)[/tex]
C. [tex]\( x = -1 \pm \sqrt{33} \)[/tex]
D. [tex]\( x = 1 \pm \sqrt{33} \)[/tex]


Sagot :

We are given the quadratic equation [tex]\( -2x^2 + x + 4 = 0 \)[/tex]. To find the solutions, we use the quadratic formula:

[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]

In our equation [tex]\( -2x^2 + x + 4 = 0 \)[/tex], the coefficients are:
[tex]\[ a = -2, \quad b = 1, \quad c = 4 \][/tex]

First, we calculate the discriminant [tex]\( \Delta \)[/tex]:

[tex]\[ \Delta = b^2 - 4ac \][/tex]

Substituting [tex]\( a \)[/tex], [tex]\( b \)[/tex], and [tex]\( c \)[/tex]:

[tex]\[ \Delta = (1)^2 - 4(-2)(4) \][/tex]
[tex]\[ \Delta = 1 + 32 \][/tex]
[tex]\[ \Delta = 33 \][/tex]

Now, we substitute back into the quadratic formula:

[tex]\[ x = \frac{-b \pm \sqrt{\Delta}}{2a} \][/tex]

Substituting [tex]\( b = 1 \)[/tex] and [tex]\( \Delta = 33 \)[/tex]:

[tex]\[ x = \frac{-1 \pm \sqrt{33}}{2(-2)} \][/tex]
[tex]\[ x = \frac{-1 \pm \sqrt{33}}{-4} \][/tex]
[tex]\[ x = \frac{-1 \pm \sqrt{33}}{-4} = \frac{1 \mp \sqrt{33}}{4} \][/tex]

Therefore, the solutions are:

[tex]\[ x = \frac{1 + \sqrt{33}}{4} \quad \text{and} \quad x = \frac{1 - \sqrt{33}}{4} \][/tex]

Comparing this to the given choices, we see that the correct answer is:

[tex]\[ \boxed{B. \, x = \frac{1 \pm \sqrt{33}}{4}} \][/tex]
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