How do you find the roots, real and imaginary, of $y=x^2 - (x-2)(2x-1)$ using the quadratic formula?

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How do you find the roots, real and imaginary, of $y=x^2 - (x-2)(2x-1)$ using the quadratic formula?

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Answer 1 · 2018-07-24T11:28:45

$x=frac{5\pm\sqrt17}{2}$

Explanation:

Given quadratic function:

$y=x^2-(x-2)(2x-1)$

$y=x^2-2x^2+5x-2$

$y=-x^2+5x-2$

The roots of given function will be at $y=0$ i.e.

$-x^2+5x-2=0$

$x^2-5x+2=0$

Now, using quadratic formula, the roots of above quadratic equation are given as

$x_{1, 2}=\frac{-(-5)\pm\sqrt{(-5)^2-4(1)(2)}}{2(1)}$

$=\frac{5\pm\sqrt17}{2}$

Thus, the roots of given quadratic function are real & distinct given as

$x=frac{5\pm\sqrt17}{2}$

Answer 2 · 2018-07-24T11:41:22

The solutions are $S={0.44, 4.56}$

Explanation:

The quadratic equation is

$y=x^2-(x-2)(2x-1)$

$=x^2-(2x^2-5x+2)$

$=x^2-2x^2+5x-2$

$=>$ , $-x^2+5x-2=0$

$a=-1$

$b=5$

$c=-2$

The discriminant is

$Delta=b^2-4ac=(5)^2-4(-1)(-2)=25-8=17$

As $Delta>0$ , there are $2$ real roots

The solution is

$x=(-b+-sqrtDelta)/(2a)$

$=(-5+-sqrt17)/(-2)$

$x_1=0.438$

$x_2=4.56$

graph{-x^2+5x-2 [-10, 10, -5, 5]}

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How do you find the roots, real and imaginary, of $y=x^2 - (x-2)(2x-1)$ using the quadratic formula?

2 Answers

Explanation:

Explanation:

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Impact of this question

Science

Math

Humanities

... and beyond

How do you find the roots, real and imaginary, of y=x^2 - (x-2)(2x-1) y=x^2 - (x-2)(2x-1) using the quadratic formula?

2 Answers

Explanation:

Explanation:

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Impact of this question

How do you find the roots, real and imaginary, of $y=x^2 - (x-2)(2x-1)$ using the quadratic formula?