Question

Find the zeroes of 15x2 – 11x + 2 and verify the relation between the zeroes and coefficients.

Answer 1

$15 {x}^{2} - 11x + 2 = 0 \\ \\ 15 {x}^{2} - 6x - 5x + 2 = 0 \\ \\ 3x(5x - 2) - 1(5x - 2) = 0 \\ \\ (3x - 1)(5x - 2) = 0 \\ \\ x = \frac{1}{3} .................x = \frac{2}{5}$

Answer 2

Solution :

We have quadratic polynomial p(x) = 15x² - 11x + 2

Zero of the polynomial p(x) = 0

$\longrightarrow\sf{15x^{2} -11x+2=0}\\\\\longrightarrow\sf{15x^{2} -5x-6x+2=0}\\\\\longrightarrow\sf{5x(3x-1)-2(3x-1)=0}\\\\\longrightarrow\sf{(3x-1)(5x-2)=0}\\\\\longrightarrow\sf{3x-1=0\:\:\:Or\:\:\:5x-2=0}\\\\\longrightarrow\sf{3x=1\:\:\:Or\:\:\:5x=2}\\\\\longrightarrow\bf{x=1/3\:\:\:Or\:\:\:x=2/5}$

∴ We get α = 1/3 & β = 2/5 are the zeroes of the polynomial .

As we know that given polynomial compared with ax² + bx + c;

• a = 15
• b = -11
• c = 2

Now;

$\underline{\mathcal{SUM\:OF\:THE\:ZEROES\::}}$

$\longrightarrow\tt{\alpha +\beta =\dfrac{-b}{a} =\bigg\lgroup\dfrac{Coefficient\:of\:x}{Coefficient\:of\:x^{2} } \bigg\rgroup}}\\\\\\\longrightarrow\tt{\dfrac{1}{3} +\dfrac{2}{5} =\dfrac{-(-11)}{15} }\\\\\\\longrightarrow\tt{\dfrac{5+6}{15} =\dfrac{11}{15} }\\\\\\\longrightarrow\bf{\dfrac{11}{15} =\dfrac{11}{15} }$

$\underline{\mathcal{PRODUCT\:OF\:THE\:ZEROES\::}}$

$\longrightarrow\tt{\alpha \times \beta =\dfrac{c}{a} =\bigg\lgroup\dfrac{Constant\:term}{Coefficient\:of\:x^{2} } \bigg\rgroup}}\\\\\\\longrightarrow\tt{\dfrac{1}{3}\times \dfrac{2}{5} =\dfrac{2}{15} }\\\\\\\longrightarrow\bf{\dfrac{2}{15} =\dfrac{2}{15} }$

Thus;

The relationship between zeroes & coefficient are verified .