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

Jan 23, 2016

$x = 0.9067 \mathmr{and} x = - 2.5734$

Explanation:

first, expand the bracket

${\left(x - 2\right)}^{2}$

$\left(x - 2\right) \left(x - 2\right)$

${x}^{2} - 4 x + 4$

then, solve the equations

$y = 4 {x}^{2} + x - 3 - \left({x}^{2} - 4 x + 4\right)$

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

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

then, by using ${b}^{2} - 4 a c$

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

where $a = 3 , b = 5 \mathmr{and} c = - 7$ into ${b}^{2} - 4 a c$

${5}^{2} - 4 \left(3\right) \left(- 7\right)$

$25 - - 84$

$109$

so, compare to this

${b}^{2} - 4 a c > 0$ : two real and different roots
${b}^{2} - 4 a c = 0$ : two real root and equals
${b}^{2} - 4 a c < 0$ : 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

$x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

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

$x = \frac{- 5 \pm \sqrt{109}}{6}$

$x = \frac{- 5 + \sqrt{109}}{6}$ and $x = \frac{- 5 - \sqrt{109}}{6}$

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

$x = 0.9067 \mathmr{and} x = - 2.5734$