Question

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Q23. Find the zeros of a quadratic polynomial given as 3x? - X-4 and also verify the relationship
between the zeroes and the coefficient​

Answer 1

GiveN :-

• A quadratic polynomial is given to us.
• 3x² - x - 4 = p(x) .

To FinD:-

• To find the zeroes of polynomials and verify the relation between the zeroes and the co-efficients .

SolutioN :-

The quadratic polynomial given to us is ,

$\sf \to \pink{ p(x) = 3x^2-x-4}$

When we Equate a quadratic polynomial with 0 , then it becomes a quadratic equation . That is ,

$\sf \to \pink{ 3x^2-x-4 = 0}$

For finding the zeroes of quadratic equation with respect to Standard form , ax² + bx + c = 0 ,

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

Now here wrt Standard form , we have ,

$\boxed{\begin{array}{c|c|c} \sf \blacksquare\:\: a = 3 &\sf \blacksquare\:\: b = (-1) &\sf \blacksquare\:\:c = (-4) \end{array}}$

$\red{\bigstar}\underline{\textsf{ Substituting the respective values , }}$

$\sf:\implies x =\dfrac{-b\pm \sqrt{b^2-4ac}}{2a} \\\\\sf:\implies x =\dfrac{-(-1) \pm \sqrt{(-1)^2-4(3)(-4)}}{2(3) } \\\\\sf:\implies x = \dfrac{1\pm \sqrt{1+48}}{6} \\\\\sf:\implies x =\dfrac{1\pm 7 }{6} \\\\\sf:\implies x =\dfrac{1+7}{6},\dfrac{1-7}{6} \\\\\sf:\implies x =\dfrac{8}{6},\dfrac{-6}{6} \\\\\sf:\implies \underset{\blue{\sf Values \ of \ x }}{\underbrace{\boxed{\pink{\frak{ x = -1 , \dfrac{4}{3} }}}}}$

$\rule{200}2$

$\red{\bigstar}\underline{\textsf{ Verification of realtionship b/w zeroes and coefficients . }}$

We know wrt Standard form the ,

$\boxed{\begin{array}{c} \textsf{\textbf{ \dag \tiny{\underline{Relationship of zeroes and coefficients of a quadratic equation}} }} \\ \\ \sf Sum \ of \ zeroes =\red{\dfrac{-b}{a}} \\ \\ \sf Product \ of \ zeroes =\red{\dfrac{c }{a}} \end{array}}$

$\sf \to Sum \ of \ zeroes \ = -1 + \dfrac{4}{3} \\\\\sf\to Sum \ of \ zeroes \ = \dfrac{-3+4}{3} \\\\\sf\to Sum \ of \ zeroes \ = \dfrac{1}{3} \\\\\sf\to Sum \ of \ zeroes \ = \dfrac{-(-1)}{3}\\\\\sf\to \underset{\purple{\sf Hence \ Proved }}{\underbrace{\boxed{\orange{\frak{ Sum \ of \ zeroes \ = \dfrac{-b}{a} }}}}}$

$\rule{200}2$

$\sf\to Product\ of \ Zeroes = (-1)\times \dfrac{4}{3} \\\\\sf\to Product\ of \ Zeroes = \dfrac{-4}{3} \\\\\sf\to \underset{\purple{\sf Hence \ Proved }}{\underbrace{\boxed{\orange{\frak{ Product \ of \ zeroes \ = \dfrac{c}{a} }}}}}$

$\rule{200}2$