In this case it is to find the roots of the polynomial.
We have then:
2x ^ 2-5x + 1 = 3
Rewriting:
2x ^ 2-5x-2 = 0
Applying resolver we have
x = (- b +/- root (b ^ 2 - 4ac)) / (2a)
Substituting values:
x = (- (- 5) +/- root ((- 5) ^ 2 - 4 (2) (- 2))) / (2 (2))
x = (- (- 5) +/- root ((25 + 16)) / (2 (2))
x = (5 +/- root (41))) / (4)
x = ((5/4) +/- (root (41)) / 4)
Answer:
x = ((5/4) +/- (root (41)) / 4)
(option 4)

8 0

3 years ago

Prove this trigonometric equation;<br><br> - tan^2x + sec^2x = 1,

alekssr [168]

Hey there :)

- tan²x + sec²x = 1 or 1 + tan²x = sec²x

sin²x + cos²x = 1
Divide the whole by cos²x

$\frac{sin^2x}{cos^2x} + \frac{cos^2x}{cos^2x} = \frac{1}{cos^2x}$

$\frac{sinx}{cosx} = tanx$ so $\frac{sin^2x}{cos^2x} = tan^2x$
and
$\frac{1}{cosx} = secx$ so $\frac{1}{cos^2x} = sec^2x$

Therefore,
tan²x + 1 = sec²x
Take tan²x to the other side {You will have the same answer}

1 = - tan²x = sec²x or sec²x - tanx = 1

3 0

2 years ago