How do you find the roots, real and imaginary, of $y=4x^2 +x -3-(x-2)^2$ using the quadratic formula?

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How do you find the roots, real and imaginary, of $y=4x^2 +x -3-(x-2)^2$ using the quadratic formula?

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Answer 1 · 2016-01-23T15:12:20

$x=0.9067 and x=-2.5734$

Explanation:

first, expand the bracket

$(x-2)^2$

$(x-2)(x-2)$

$x^2-4x+4$

then, solve the equations

$y=4x^2+ x-3-(x^2-4x+4)$

$y=4x^2+ x-3-x^2+4x-4$

$y=3x^2+ 5x-7$

then, by using $b^2-4ac$

for the equation: $y=3x^2+ 5x-7$

where $a=3, b=5 and c=-7$ into $b^2-4ac$

$5^2-4(3)(-7)$

$25--84$

$109$

so, compare to this

$b^2-4ac>0$ : two real and different roots
$b^2-4ac=0$ : two real root and equals
$b^2-4ac<0$ : no real roots or (the roots are complexes)

so, $109>0$ means two real and different roots

thus, you must use this formula to find the imaginary roots

$x= (-b+- sqrt(b^2-4ac ))/ (2a)$

$x= (-5+- sqrt(5^2-4(3)(-7) ))/ (2(3)$

$x= (-5+- sqrt(109))/ 6$

$x= (-5+sqrt(109))/ 6$ and $x= (-5- sqrt(109))/ 6$

solve it and u will get the values of x which is

$x=0.9067 and x=-2.5734$

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How do you find the roots, real and imaginary, of $y=4x^2 +x -3-(x-2)^2$ using the quadratic formula?

1 Answer

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=4x^2 +x -3-(x-2)^2 y=4x^2 +x -3-(x-2)^2 using the quadratic formula?

1 Answer

Explanation:

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Impact of this question

How do you find the roots, real and imaginary, of $y=4x^2 +x -3-(x-2)^2$ using the quadratic formula?