Question

use the quadratic formula to find the zeros of the function round to the tense of necessary y= 17x^2-21

Answer 1

$y=17x^2-21\\\\\text{The zeros of the function:}\ y=0\iff17x^2-21=0\ \ \ |+21\\\\17x^2=21\ \ \ |:17\\\\x^2=\dfrac{21}{17}\to x=\pm\sqrt{\dfrax{21}{17}}\\\\x\approx-1.1\ \vee\ x\approx1.1$

$\text{Use the quadratic formula:}\\\\y=17x^2-21\\\\a=17;\ b=0;\ c=-21\\\\b^2-4ac=0^2-4\cdot17\cdot(-21)=1428\\\\x_1=\dfrac{-b-\sqrt{b^2-4ac}}{2a};\ x_2=\dfrac{-b+\sqrt{b^2-4ac}}{2a}\\\\x_1=\dfrac{-0-\sqrt{1428}}{2\cdot17}=-\dfrac{\sqrt{1428}}{34}\approx-1.1\\\\x_2=\dfrac{-0+\sqrt{1428}}{2\cdot17}=\dfrac{\sqrt{1428}}{34}\approx1.1$