How do you find the roots, real and imaginary, of $y = x^{2} - 3 x + 2$ $y=x^2 - 3x + 2$ using the quadratic formula?

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How do you find the roots, real and imaginary, of $y = x^{2} - 3 x + 2$ $y=x^2 - 3x + 2$ using the quadratic formula?

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Answer 1 · 2015-12-09T04:34:46

You can find all the roots of $y = x^{2} - 3 x + 2$ $y = x^2 - 3x + 2$ by "plugging in" values from the equation into the quadratic formula to find that the roots are $(2, 0)$ $(2, 0)$ and $(1, 0)$ $(1, 0)$ .

Explanation:

The quadratic formula is:

$x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$ $x = (-b +- sqrt(b^2 - 4ac))/(2a)$

The equation representing any given quadratic, or parabola, can be:

$y = a x^{2} + b x + c$ $y = ax^2 + bx + c$

So all we need to do is assign values of our equation to their respective variables, and then substitute them into the quadratic formula to solve. In the equation $y = x^{2} - 3 x + 2$ $y = x^2 - 3x + 2$ :

$a = 1$ $a = 1$ , $b = (- 3)$ $b = (-3)$ , and $c = 2$ $c = 2$

$x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$ $x = (-b +- sqrt(b^2 - 4ac))/(2a)$ Given Quadratic Formula

$x = \frac{- (- 3) \pm \sqrt{{(- 3)}^{2} - 4 (1) (2)}}{2 (1)}$ $x = (-(-3) +- sqrt((-3)^2 - 4(1)(2)))/(2(1))$ Substitute

$x = \frac{3 \pm \sqrt{9 - 4 (2)}}{2}$ $x = (3 +- sqrt(9 - 4(2)))/(2)$ Simplify

$x = \frac{3 \pm \sqrt{9 - 8}}{2}$ $x = (3 +- sqrt(9 - 8))/(2)$ Simplify Some More

$x = \frac{3 \pm \sqrt{1}}{2}$ $x = (3 +- sqrt(1))/(2)$ Simplify A Little More

$x = \frac{3 \pm 1}{2}$ $x = (3 +- 1)/(2)$ Simplify Even More

$x = \frac{3 + 1}{2}$ $x = (3 + 1)/2$ , OR $x = \frac{3 - 1}{2}$ $x = (3 - 1)/2$ Isolate Both Possible Roots

$x = 2$ $x = 2$ OR $x = 1$ $x = 1$ Simplify (For the Last Time..)

This means that the roots, or x-intercepts, of our quadratic equation are at the points: $(2, 0)$ $(2, 0)$ and $(1, 0)$ $(1, 0)$ .

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How do you find the roots, real and imaginary, of $y = x^{2} - 3 x + 2$ $y=x^2 - 3x + 2$ using the quadratic formula?

1 Answer

Explanation:

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Science

Math

Humanities

... and beyond

How do you find the roots, real and imaginary, of y=x^2 - 3x + 2 y=x2−3x+2y=x^2 - 3x + 2 using the quadratic formula?

1 Answer

Explanation:

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How do you find the roots, real and imaginary, of $y = x^{2} - 3 x + 2$ $y=x^2 - 3x + 2$ using the quadratic formula?