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Roots of Quadratic Equation

Find the zero...

Question

Find the zeros of the following quadratic polynomials
(i) $x^{2} - 2 x - 8$
(ii) $4 s^{2} - 4 s + 1$

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Solution

i) Given polynomial $x^{2} - 2 x - 8$
Here, the middle term $- 2 x$ expressed as sum of two terms $4 x$ and $- 2 x$ such that its product $4 x \times (- 2 x) = - 8 x^{2}$ is equals to the product of extreme terms.
[ $x^{2} \times (- 8) = - 8 x^{2}$ ]
Thus, $x^{2} - 4 x + 2 x - 8 = 0 x (x - 4) + 2 (x - 4) = 0$
$(x + 2) (x - 4) = 0 x + 2 = 0$ and $x - 4 = 0$
$∴ x = - 2, 4$
ii) Given polynomial $4 s^{2} - 4 s + 1$
Here, the middle term is $- 4 s$ is sum of $- 2 s$ and $- 2 s$ such that its product is $- 2 s \times - 2 s = 4 s^{2}$ is equal to product of extreme terms.
$4 s^{2} - 2 s - 2 s + 1 = 0$
$2 s (2 s - 1) - 1 (2 s - 1) = 0$
$(2 s - 1) (2 s - 1) =$
$(2 s - 1)^{2} = 0$
$2 s - 1 = 0$ [Taking square root on both sides]
$2 s = 1$
$∴ s = \frac{1}{2}$

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Similar questions

Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients.
(i) $x^{2} - 2 x - 8$

(ii) $4 s^{2} - 4 s + 1$

(iii) $6 x^{2} - 3 - 7 x$
(iv) $4 u^{2} + 8 u$

(v) $t^{2} - 15$

(vi) $3 x^{2} - x - 4$

f (x) = x^{2} - 2 x - 8

g (x) = 4 s^{2} - 4 s + 1

f (x) = x^{2} - (\sqrt{3} + 1) x + \sqrt{3}

x^{2} - 3 - 7 x

p (x) = x^{2} + 2 \sqrt{2} x - 6

q (x) = \sqrt{3} x^{2} + 10 x + 7 \sqrt{3}

g (x) = a (x^{2} + 1) - x (a^{2} + 1)

Question 1 (ii)
Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients.

$4 s^{2} - 4 s + 1$

f (x) = x^{2} - 2 x - 8

g (s), = 4 s^{2} - 4 s + 1

h (t) = t^{2} - 15

6 x^{2} - 3 - 7 x

p (x) = x^{2} + 2 \sqrt{2} - 6

q (x) = \sqrt{3 x^{2}} + 10 x + \sqrt[7]{3}

f (x) = x^{2} - (\sqrt{3} + 1) x + \sqrt{3}

g (x) = a (x^{2} + 1) - x (a^{2} + 1)

4 s^{2} - 4 s + 1

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