How do you find the x intercepts for $y = - 2 x^{2} + 5 x - 1$ $y = -2x^2 + 5x - 1$ ?

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How do you find the x intercepts for $y = - 2 x^{2} + 5 x - 1$ $y = -2x^2 + 5x - 1$ ?

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Answer 1 · 2015-06-30T23:15:17

Solve $y = - 2 x^{2} + 5 x - 1 = 0$ $y = -2x^2 + 5x - 1 = 0$

Explanation:

$D = d^{2} = 25 - 8 = 17 \to d = \pm \sqrt{17}$ $D = d^2 = 25 - 8 = 17 -> d = +- sqrt17$

$x = \frac{5}{4} \pm \frac{\sqrt{17}}{4}$ $x = 5/4 +- (sqrt17)/4$

Answer 2 · 2015-07-01T02:10:10

$x = \frac{5}{4} \pm \frac{\sqrt{17}}{4}$ $x = 5/4 +-sqrt(17)/4$

Explanation:

This answer is the same as given by Nghi N. I am simply providing a more detailed explanation.

Since the x-intercepts occur when $y = 0$ $y=0$ we solve the given equation with $0$ $0$ in place of $y$ $y$ to find the x-intercepts.

$- 2 x^{2} + 5 x - 1 = 0$ $−2x^2+5x−1=0$

Any parabolic equation of the form $a x^{2} + b x + c = 0$ $ax^2+bx+c=0$
can be solved using the quadratic formula:
$XXXX$ $color(white)("XXXX")$ $x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$ $x = (-b+-sqrt(b^2-4ac))/(2a)$

Applying the values from our example:
$XXXX$ $color(white)("XXXX")$ $x = \frac{- 5 \pm \sqrt{5^{2} - 4 (- 2) (- 1)}}{2 (- 2)}$ $x = (-5+-sqrt(5^2-4(-2)(-1)))/(2(-2))$

$XXXX$ $color(white)("XXXX")$ $x = \frac{- 5 \pm \sqrt{25 - 8}}{- 4}$ $x = (-5 +- sqrt(25-8))/(-4)$

$XXXX$ $color(white)("XXXX")$ $x = \frac{5}{4} \pm \frac{\sqrt{17}}{4}$ $x = 5/4 +-sqrt(17)/4$

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How do you find the x intercepts for $y = - 2 x^{2} + 5 x - 1$ $y = -2x^2 + 5x - 1$ ?

2 Answers

Explanation:

Explanation:

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Science

Math

Humanities

... and beyond

How do you find the x intercepts for y=−2x2+5x−1y = -2x^2 + 5x - 1?

2 Answers

Explanation:

Explanation:

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Impact of this question

How do you find the x intercepts for $y = - 2 x^{2} + 5 x - 1$ $y = -2x^2 + 5x - 1$ ?