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# Find the zeroes of the following quadratic polynomials and verify the relationship between the zeros and the coefficients. (i) $$x^2 - 2x - 8$$ (ii) $$4s^2 - 4s + 1$$ (iii) $$6x^2 - 3 - 7x$$ (iv) $$4u^2 + 8u$$ (v) $$t^2 - 15$$ (vi) $$3x^2 - x - 4$$

General form of equation is $$ax^2 + bx + c$$
where, a = Coefficient of $$x^2$$
b = Coefficient of x
c = Constant term

Once we'll find the zeroes of the polynomial we can verify it using the following rules

Sum of zeroes = $${{-b}\over {a}}$$
Product of zeroes = $${{c}\over {a}}$$

(i) $$x^2 - 2x - 8$$
=$$x^2 - 4x + 2x - 8$$
= $$x(x - 4) + 2(x - 4)$$
= $$(x + 2)(x - 4)$$

So, zeroes are -2 and 4
Now, Verification:

Sum of the zeroes = 2
= $${{-b}\over {a}}$$ = $${{-(-2)}\over {1}}$$ = 2
Product of the zeroes = -8
= $${{c}\over {a}}$$ = $${{-8}\over {1}}$$ = -8

(ii) $$4s^2 - 4s + 1$$
=$$4s^2 - 2s - 2s + 18$$
= $$2s(2s - 1) - 1(2s - 1)$$
= $$(2s - 1)(2s - 1)$$

So, zeroes are $${{1}\over {2}}$$ and $${{1}\over {2}}$$
Now, Verification:

Sum of the zeroes = 1
= $${{-b}\over {a}}$$ = $${{-(-4)}\over {4}}$$ = 1
Product of the zeroes = $${{1}\over {4}}$$
= $${{c}\over {a}}$$ = $${{1}\over {4}}$$

(iii) $$6x^2 - 7x - 3$$
=$$6x^2 - 9x + 2x -3$$
= $$3x(2x - 3) + 1(2x - 3)$$
= $$(3x + 1)(2x - 3)$$

So, zeroes are $${{-1}\over {3}}$$ and $${{3}\over {2}}$$
Now, Verification:
Sum of the zeroes
= $${{-1}\over {3}} + {{3}\over {2}}$$ = $${{7}\over {6}}$$
= $${{-b}\over {a}}$$ = $${{-(-7)}\over {6}}$$
Product of the zeroes
= $${{-1}\over {3}} × {{3}\over {2}}$$ = $${{-1}\over {2}}$$
= $${{c}\over {a}}$$ = $${{-3}\over {6}}$$ = $${{-1}\over {2}}$$

(iv) $$4u^2 + 8u$$
=$$4u(u + 2)$$
=$$4u = 0$$
=$$u = 0$$
=$$u = -2$$

So, zeroes are 0 and -2
Now, Verification:

Sum of the zeroes = -2
= $${{-b}\over {a}}$$ = $${{-(8)}\over {4}}$$ = -2
Product of the zeroes = 0
= $${{c}\over {a}}$$ = $${{0}\over {4}}$$ = 0

(v) $$t^2 - 15$$
= $$t^2 - (\sqrt{15})^2$$
= $$(t + \sqrt{15})(t -\sqrt{15})$$

So, zeroes are $$-\sqrt{15}$$ and $$\sqrt{15}$$
Now, Verification:
Sum of the zeroes = 0
= $${{-b}\over {a}}$$ = $${{-(0)}\over {1}}$$ = 0
Product of the zeroes = -15
= $${{c}\over {a}}$$ = $${{-15}\over {1}}$$ = -15

(vi) $$3x^2 - x - 4$$
=$$3x^2 - 4x + 3x -4$$
= $$x(3x - 4) + 1(3x - 4)$$
= $$(3x - 4)(x + 1)$$

So, zeroes are $${{4}\over {3}}$$ and -1
Now, Verification:
Sum of the zeroes
= $${{4}\over {3}} + (-1)$$ = $${{1}\over {3}}$$
= $${{-b}\over {a}}$$ = $${{-(-1)}\over {3}}$$ = $${{1}\over {3}}$$
Product of the zeroes
= $${{4}\over {3}} × (-1)$$ = $${{-4}\over {3}}$$
= $${{c}\over {a}}$$ = $${{-4}\over {3}}$$