Question

Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients:
(i) x² + 10x + 16
(ii) x² $-$ 4x $-$ 5

Answer 1

$$\begin{gathered} \left(\mathrm{i}\right)\mathrm{We}\mathrm{have}{x}^2+10x+16=x^2+8x+2x+16 \\ =x\left(x+8\right)+2\left(x+8\right) \\ =\left(x+8\right)\left(x+2\right) \\ \mathrm{The}\mathrm{polynomial}\mathrm{has}\mathrm{zeroes}\mathrm{when}p\left(x\right)=0 \\ \mathrm{i}.\mathrm{e}.,x+8=0\mathrm{or}x+2=0 \\ \Rightarrow x=-8\mathrm{or}x=-2 \\ \therefore \mathrm{The}\mathrm{zeroes}\mathrm{of}\mathrm{the}\mathrm{polynomial}{x}^2+10x+16\mathrm{are}-8\mathrm{and}-2. \\ \mathrm{Now},\mathrm{sum}\mathrm{of}\mathrm{the}\mathrm{zeroes}=\left(-8\right)+\left(-2\right)=-10=\frac{-\left(10\right)}{1}=\frac{-b}{a} \\ \mathrm{and}\mathrm{product}\mathrm{of}\mathrm{the}\mathrm{zeroes}=\left(-8\right)\times \left(-2\right)=16=\frac{16}{1}=\frac{c}{a} \\ \mathrm{Hence},\mathrm{verified}. \end{gathered}$$

$$\begin{gathered} \left(\mathrm{ii}\right)\mathrm{We}\mathrm{have}{x}^2-4x-5=x^2-5x+x-5 \\ =x\left(x-5\right)+1\left(x-5\right) \\ =\left(x-5\right)\left(x+1\right) \\ \mathrm{The}\mathrm{polynomial}\mathrm{has}\mathrm{zeroes}\mathrm{when}p\left(x\right)=0 \\ \mathrm{i}.\mathrm{e}.,x-5=0\mathrm{or}x+1=0 \\ \Rightarrow x=5\mathrm{or}x=-1 \\ \therefore \mathrm{The}\mathrm{zeroes}\mathrm{of}\mathrm{the}\mathrm{polynomial}{x}^2+10x+16\mathrm{are}5\mathrm{and}-1. \\ \mathrm{Now},\mathrm{sum}\mathrm{of}\mathrm{the}\mathrm{zeroes}=5+\left(-1\right)=4=\frac{-\left(-4\right)}{1}=\frac{-b}{a} \\ \mathrm{and}\mathrm{product}\mathrm{of}\mathrm{the}\mathrm{zeroes}=5\times \left(-1\right)=-5=\frac{-5}{1}=\frac{c}{a} \\ \mathrm{Hence},\mathrm{verified}. \end{gathered}$$