# Find the zeroes of the quadratic polynomials and verify a relationship between zeroes and its coefficients.

Question : Find the zeroes of the quadratic polynomials and verify a relationship between zeroes and its 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)  $$3x^2 – x – 4$$

Solution :

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

So, the value of $$x^2 – 2x – 8$$ is zero when x – 4 = 0 or x + 2 = 0. i.e. when x = 4 or x = -2.

So, the zeroes of $$x^2 – 2x – 8$$ are 4, -2.

Sum of the zeroes = 4 – 2 = 2 = $$-(-2)\over 1$$ = $$-coefficient of x\over coefficient of x^2$$ = 2

Product of the zeroes = 4(-2) = -8 = $$-8\over 1$$ = $$constant term\over coefficient of x^2$$ = -8

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

So, the value of $$4s^2 – 4s + 1$$ is zero when 2s – 1 = 0 or s = $$1\over 2$$

So, the zeroes of $$x^2 – 2x – 8x$$ are $$1\over 2$$, $$1\over 2$$

Sum of the zeroes = $$1\over 2$$ + $$1\over 2$$ = 1 = $$-(-4)\over 4$$ = $$-coefficient of s\over coefficient of s^2$$ = 1

Product of the zeroes = ($$1\over 2$$)($$1\over 2$$) = $$1\over 4$$ = $$1\over 4$$ = $$constant term\over coefficient of s^2$$ = $$1\over 4$$

(iii) $$6x^2 – 3 – 7x$$ = $$6x^2 – 7x – 3$$ = $$6x^2 – 9x – 2x – 3$$

= 3x(2x – 3) + 1(2x – 3) = (3x + 1)(2x – 3)

So, the value of $$6x^2 – 3 – 7x$$ is zero when the value of (3x + 1)(2x – 3) = 0 i.e.

when 3x + 1 = 0 or 2x – 3 = 0  i.e. when x = $$-1\over 3$$ or x = $$3\over 2$$.

So, the zeroes of $$6x^2 – 3 – 7x$$ are $$-1\over 3$$ and $$3\over 2$$.

Sum of the zeroes = $$-1\over 3$$ + $$3\over 2$$ = $$7\over 6$$ = $$-(-7)\over 6$$ = $$-coefficient of x\over coefficient of x^2$$ = $$7\over 6$$

Product of the zeroes = ($$-1\over 3$$)($$3\over 2$$) = $$-3\over 6$$ = ($$-1\over 3$$)($$3\over 2$$) = $$constant term\over coefficient of x^2$$ = $$-3\over 6$$

(iv) We have : $$4u^2 + 8u$$ = 4u(u + 2)

The value of $$4u^2 + 8u$$ is zero when the value of 4u(u + 2) = 0 i.e. when u = 0 or u + 2 = 0 i.e. u = 0 or u = -2.

So, the zeroes of $$4u^2 + 8u$$ are 0, -2.

Sum of the zeroes = 0 + (- 2) = -2 = $$-(-8)\over 4$$ = $$-coefficient of u\over coefficient of u^2$$ = –2

Product of the zeroes = (0)(-2) = 0 = $$0\over 4$$ = $$constant term\over coefficient of u^2$$ = 0

(v)  We have, $$t^2 – 15$$ = $$(t – \sqrt{15})$$$$(t + \sqrt{15})$$

The value of $$t^2 – 15$$ is zero when the value of $$(t – \sqrt{15})$$$$(t + \sqrt{15})$$ is 0, i.e. when $$(t – \sqrt{15})$$ = 0 or $$(t + \sqrt{15})$$ = 0

i.e. when t = $$\sqrt{15}$$, $$-\sqrt{15}$$.

So, the zeroes of $$t^2 – 15$$ are  $$\sqrt{15}$$, $$-\sqrt{15}$$.

Sum of the zeroes = ($$\sqrt{15}$$) +  ($$-\sqrt{15}$$) = 0 = $$-(-0)\over 1$$ = $$-coefficient of t\over coefficient of t^2$$ = 0

Product of the zeroes = ($$\sqrt{15}$$)($$-\sqrt{15}$$) = -15 = $$-15\over 1$$ = $$constant term\over coefficient of t^2$$ = -15

(vi)  We have : $$3x^2 – x – 4$$ = $$3x^2 – 3x – 4x – 4$$ = 3x(x +1) – 4(x + 1) = (x + 1)(3x – 4)

The value of $$3x^2 – x – 4$$ is zero when the value of (x + 1)(3x – 4) is 0, i.e. when x + 1 = 0 or 3x – 4 = 0, i.e. when x = -1 or x = $$4\over 3$$

So, the zeroes of $$3x^2 – x – 4$$ are -1, $$4\over 3$$.

Sum of the zeroes = -1 + $$4\over 3$$ = $$-3 + 4\over 3$$ = $$1\over 3$$ = $$-(-1)\over 3$$ =  $$-coefficient of x\over coefficient of x^2$$ = $$1\over 3$$.

Product of the zeroes = (-1)($$4\over 3$$) = $$constant term\over coefficient of x^2$$ = $$-4\over 3$$