Question

If the y-coordinate of the vertex of the graph of y = ax2 + bx + c, a < 0 is zero, then the zero of the quadratic polynomial is equal to

Answer 1

$\text{Given polynomial is}\;y=a\,x^2+b\,x+c$

$\text{First we make the R.H.S of the polynomial equation as a perfect square}$

$y=a\,x^2+b\,x+c$

$y=a(x^2+\frac{b}{a}x+\frac{c}{a})$

$y=a[(x^2+\frac{b}{a}x+\frac{b^2}{4a^2})-\frac{b^2}{4a^2}+\frac{c}{a}]$

$y=a[(x-\frac{b}{2a})^2-\frac{b^2}{4a^2}+\frac{c}{a}]$

$y=a(x-\frac{b}{2a})^2+\frac{4ac-b^2}{4a}$

$y-(\frac{4ac-b^2}{4a})=a(x-\frac{b}{2a})^2$

$(x-\frac{b}{2a})^2=\frac{1}{a}(y-(\frac{4ac-b^2}{4a}))$

$\text{This equation is of the form}\;(x-h)^2=4a(y-k)$

$\implies\,\text{Vertex}\,(h,k)=(\frac{b}{2a},\frac{4ac-b^2}{4a})$

$\text{Since the y coordinate of the vertex is 0, we have}$

$\frac{4ac-b^2}{4a}=0$

$\implies\,b^2-4ac=0$

$\implies\text{Zeros of the quadratic polynomial ax2 + bx + c are equal}$

$\text{Then, its zeros are found by using}$

$x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$

$x=\dfrac{-b\pm\sqrt{0}}{2a}$

$\implies\bf\,x=\dfrac{-b}{2a}$

$\therefore\textbf{Zeros of the given polynomial is \dfrac{-b}{2a} }$