# Find a quadratic polynomial whose sum of zeroes and product of zeroes are respectively.

Question : Find a quadratic polynomial whose sum of zeroes and product of zeroes are respectively.

(i)  $$1\over 4$$, -1

(ii)  $$\sqrt{2}$$, $$1\over 3$$

(iii)  0, $$\sqrt{5}$$

(iv)  1, 1

(v)  $$-1\over 4$$, $$1\over 4$$

(vi)  4, 1

Solution : Let the polynomial be $$ax^2 + bx + c$$ and its zeroes be $$\alpha$$ and $$\beta$$.

(i)  Here, $$\alpha$$ + $$\beta$$ = $$1\over 4$$  and $$\alpha$$.$$\beta$$ = -1

Thus, the polynomial formed = $$x^2$$ – (sum of zeroes) x + Product of zeroes

= $$x^2$$ – ($$1\over 4$$)x – 1

= $$4x^2 – x – 4$$

(ii)  Here, $$\alpha$$ + $$\beta$$ = $$\sqrt{2}$$  and $$\alpha$$.$$\beta$$ = $$1\over 3$$

Thus, the polynomial formed = $$x^2$$ – (sum of zeroes) x + Product of zeroes

= $$x^2$$ – ($$\sqrt{2}$$)x + $$1\over 3$$

= $$3x^2 – 3\sqrt{2}x + 1$$

(iii)  Here, $$\alpha$$ + $$\beta$$ = 0  and $$\alpha$$.$$\beta$$ = $$\sqrt{5}$$

Thus, the polynomial formed = $$x^2$$ – (sum of zeroes) x + Product of zeroes

= $$x^2$$ – (0)x + $$\sqrt{5}$$

= $$x^2 + \sqrt{5}$$

(iv)  Here, $$\alpha$$ + $$\beta$$ = 1  and $$\alpha$$.$$\beta$$ = 1

Thus, the polynomial formed = $$x^2$$ – (sum of zeroes) x + Product of zeroes

= $$x^2$$ – (1)x + 1

= $$x^2 – x + 1$$

(v)  Here, $$\alpha$$ + $$\beta$$ = $$-1\over 4$$  and $$\alpha$$.$$\beta$$ = $$1\over 4$$

Thus, the polynomial formed = $$x^2$$ – (sum of zeroes) x + Product of zeroes

= $$x^2$$ – ($$-1\over 4$$)x + $$1\over 4$$

= $$4x^2 + x + 1$$

(vi)  Here, $$\alpha$$ + $$\beta$$ = 4  and $$\alpha$$.$$\beta$$ = 1

Thus, the polynomial formed = $$x^2$$ – (sum of zeroes) x + Product of zeroes

= $$x^2$$ – (4)x + 1

= $$x^2 – 4x + 1$$