Question

Show that $4$ an $1$ are the zeroes of quadratic polynomial $p\left(x\right)=x^2-5x+4$. Also verify the relation between the zeroes and the coefficients of the polynomial.

Answer 1

$p\left(x\right)=x^2-5x+4$

To show: $4$ and $1$ are zeroes of $p\left(x\right)$

Proof: $p\left(4\right)=(4{)}^2-5(4)+4$

$p\left(4\right)=16-20+4$

$p\left(4\right)=20-20=0$

$p\left(1\right)=1^2-5\left(1\right)+4$

$p\left(1\right)=1-5+4$

$p\left(1\right)=5-5$

$p\left(1\right)=0$

$\because p\left(4\right)$ & $p\left(1\right)$ is equal to zero

$\therefore 4$ & $1$ are the zeroes of $p\left(x\right)$

For polynomial $a{x}^2+bx+c$ having roots $\alpha$ & $\beta$

Sum of the roots $=\alpha +\beta =\frac{-b}{a}$

Product of the roots $=\alpha \beta =\frac{c}{a}$

Sum of zeroes of $p\left(x\right)=4+1=5$

using above formula,

Sum of zeroes $=\frac{-(-5)}{1}=5$

Product of zeroes of $p\left(x\right)=4\times 1=4$

Using above formula,

Product of zeroes $=\frac{4}{1}=4$

Hence, verified.