Question

4x^2+13x+3
find the zeroes of the following quadratic polynomial and verify the relationship between the zeroes and coefficient​
plz solve​

Answer 1

S O L U T I O N :

We have p(x) = 4x² + 13x + 3

Zero of the polynomial p(x) = 0

Using quadratic formula :

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

• a = 4
• b = 13
• c = 3

So;

$\longrightarrow\tt{x=\dfrac{-b\pm\sqrt{b^{2}-4ac } }{2a} }\\\\\\\longrightarrow\tt{x=\dfrac{-13\pm\sqrt{(13)^{2} -4\times 4\times 3} }{2\times 4} }\\\\\\\longrightarrow\tt{x=\dfrac{-13\pm\sqrt{169-48} }{8} }\\\\\\\longrightarrow\tt{x=\dfrac{-13\pm\sqrt{121} }{8} }\\\\\\\longrightarrow\tt{x=\dfrac{-13\pm11}{8} }\\\\\\\longrightarrow\tt{x=\dfrac{-13+11}{8} \:\:Or\:\:x=\dfrac{-13-11}{8} }\\\\\\\longrightarrow\tt{x=\cancel{\dfrac{-2}{8}} \:\:Or\:\:x=\cancel{\dfrac{-24}{8} }}$

$\longrightarrow\bf{x=-\dfrac{1}{4} \:\:\:Or\:\:\:x=-3}$

∴ The α = -1/4 and β = -3 are the zeroes of the polynomial.

Now;

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

$\longrightarrow\tt{\alpha +\beta =\dfrac{-b} {a}=\dfrac{Coefficient\:of\:x}{Coefficient\:of\:x^{2} } }\\\\\\\longrightarrow\tt{-\dfrac{1}{4} +(-3)=\dfrac{-13}{4} }\\\\\\\longrightarrow\tt{-\dfrac{1}{4}-3=\dfrac{-13}{4}}\\\\\\\longrightarrow\tt{\dfrac{-1-12}{4} =\dfrac{-13}{4}}\\\\\\\longrightarrow\bf{\dfrac{-13}{4} =\dfrac{-13}{4} }$

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

$\longrightarrow\tt{\alpha \times \beta =\dfrac{c} {a}=\dfrac{Constant\:term}{Coefficient\:of\:x^{2} } }\\\\\\\longrightarrow\tt{-\dfrac{1}{4} \times (-3)=\dfrac{3}{4} }\\\\\\\ \longrightarrow\bf{\dfrac{3}{4} =\dfrac{3}{4} }$

Thus;

Relationship between zeroes and coefficient is verified .

Answer 2

Given ,

The polynomial is 4(x)² + 13x + 3

By middle term splitting method ,

$\hookrightarrow 4 {(x)}^{2} + 13x + 3 = 0 \\ \\ \hookrightarrow</p><p>4{(x)}^{2}+ 12x + x + 3 = 0 \\ \\ \hookrightarrow</p><p>4x(x + 3) + 1(x + 3) = 0 \\ \\ \hookrightarrow</p><p>(4x + 1)(x + 3) = 0 \\ \\ \hookrightarrow</p><p>x = - \frac{1}{4} \: \: or \: \: x = - 3$

$\therefore \sf \underline{The \: zeroes \: of \: polynomial \: are - \frac{1}{4} \: and - 3 }$

We know that , the relationship between zeroes and coefficient of polynomial is given by

$\star \: \sf \fbox{ \alpha + \beta = - \frac{b}{a} }$

Thus ,

$\mapsto \frac{ - 1}{4} +( - 3) = - \frac{13}{4} \\ \\ \mapsto - \frac{ 13}{4} = - \frac{13}{4}$

And

$\star \: \sf \fbox{ \alpha \times \beta = \frac{c}{a}}$

Thus ,

$\mapsto \frac{ - 1}{4} \times ( - 3) = \frac{3}{4} \\ \\ \mapsto \frac{3}{4} = \frac{3}{4}$

Hence verified