How do you solve $11 z^{2} - z = 3$ $11z^2-z=3$ using the quadratic formula?

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Answer 1 · 2016-12-24T05:25:53

$z = \frac{1 + \sqrt{133}}{22} and \frac{1 - \sqrt{33}}{22}$ $z = [1 + sqrt(133)]/22 and [1 - sqrt(33)]/22$

$z \approx \pm 0.5697$ $z~~ +-0.5697$

The general form of quadratic equation is $a x^{2} + b x + c = 0$ $ax^2 +bx +c = 0$

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

Here, the equation is $11 z^{2} - z = 3$ $11z^2 - z = 3$

or, $11 z^{2} - z - 3 = 0$ $" "11z^2 - z -3 = 0$

comparing this with $a x^{2} + b x + c = 0$ $ax^2 +bx +c = 0$

we get $x = z, a = 11, b = - 1 and c = - 3$ $x =z, a = 11, b = - 1 and c = - 3$

Now, $z = = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$ $z = =(-b +-sqrt(b^2 - 4ac))/(2a)$ put values of a,b and c

or, $z = \frac{- (- 1) \pm \sqrt{{(- 1)}^{2} - 4 (11) (- 3)}}{2 (11)}$ $z =(-(-1) +-sqrt((-1)^2 - 4(11)(-3)))/(2(11))$

or, $z = = \frac{1 \pm \sqrt{1 + 132}}{22}$ $z = =(1 +-sqrt(1 +132))/(22)$

or, $z = \frac{1 + \sqrt{133}}{22}$ $z = [1 + sqrt{133}]/22$ and $z = \frac{1 - \sqrt{133}}{22}$ $z = (1 - sqrt{133})/22$

How do you solve 11z2−z=311z^2-z=3 using the quadratic formula?