Question

Find the quadratic polynomial, for the zeroes a, B given in each case

(1) 1/2,3/2​

Answer 1

5

Step-by-step explanation:

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Answer 2

$\large\sf\underline{Answer:-}$

• $\sf\:⟹\:x^{2}-2x+\frac{3}{4}=0$

$\large\sf\underline{Given:-}$

$\sf\:Two\:roots\:or\:zeroes\:of\:the\:polynomial$

• $\sf\:\frac{1}{2}\:and\:\frac{3}{2}$

$\large\sf\underline{To\:find:-}$

• $\sf\:The\:quadratic\:polynomial$

$\large\sf\underline{Solution :-}$

We know :

Formula for quadratic equation is -

$\tt\red{\:x^{2}-(sum\:of\:the\:roots)x+(product\:of\:the\:roots)=0}$

So, in our case :

$\sf\:Sum\:of\:the\:roots$

$\sf↦\:\frac{1}{2}+\frac{3}{2}$

$\sf↦\:\frac{1+3}{2}$

$\sf↦\:\frac{4}{2}$

$\tt\purple{↦Sum\:of\:the\:roots=\:2}$

$\sf\:Product\:of\:the\:roots$

$\sf↦\:\frac{1}{2} \times \frac{3}{2}$

$\sf↦\:\frac{3}{4}$

$\tt\purple{↦Product\:of\:the\:roots=\:\frac{3}{4}}$

Now substituting the value in our formula :

$\tt\red{\:x^{2}-(sum\:of\:the\:roots)x+(product\:of\:the\:roots)=0}$

$\sf⟹\:x^{2}-(2)x+(\frac{3}{4})=0$

$\tt\red{⟹\:x^{2}-2x+\frac{3}{4}=0}$

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