Question

Questions:
Find the roots of the equation x2 + x - (a + 2) (a + 1) = 0.
Solve the following quadratic equations for x and give your answer correct to 2 significant figures:
i) 4x2-5x-3=0
ii)x-18/x=6
Solve by using quadratic formula:
2x square+√5x-5=0

Please solve these sums quickly
Its urgent...​

Answer 1

(i) using factorization :

=> x² + x - (a + 1)(a + 2)

Split the middle term(with x) in such a manner so that product of parts is equal to product of coeff. of x² and constant. Here, parts are (a + 2) and (a + 1).

=> x² + ((a + 2) - (a + 1))x - (a + 1)(a + 2) = 0

=> x² + (a + 2)x - (a + 1)x - (a + 1)(a + 2) = 0

=> x(x + a + 2) - (a + 1)(x + a + 2) = 0

=> (x + a + 2)(x - (a + 1)) = 0

=> x = - (a + 2) or a + 1.

(ii) 4x² - 5x - 3 = 0

Using quadraction formula :

x = [-(-5) ± √(-5)² - 4(4)(-3)) ]/2(4)

x = (5 ± √73)/8

x = (5 ± 8.42)/8

x = (5 - 8.42)/8 or (5 + 8.42)/8

x = - 0.44 or 1.69

(iii) x - 18/x = 6

=> (x² - 18)/x = 6 => x² - 6x - 18 = 0

Using quadratic formula :

x = [-(-6) ± √(-6)² - 4(1)(-18)) ]/2(1)

x = (6 ± √108)/2 = (6 ± 6√3)/2

x = 3 + 3√3 or 3 - 3√2

x = 8.19 or - 2.19

(iv) 2x² + √5x - 5 = 0

Using quadratic formula :

=> x = [-√5 ± √(√5)² - 4(2)(-5)) ]/2(2)

=> x = [-√5 ± √(5 + 40) ]/4

=> x = (-√5 ± √45)/4

=> x = (-√5 ± 3√5)/4

=> x = (-√5 + 3√5)/4 of (-√5 - 3√5)/4

=> x = 2√5/4 or -4√5/4

=> x = √(5)/2 or - √5

Answer 2

Required Answer :-

1]

$\sf x = \dfrac{-1\pm \sqrt{(1)^2 - 4 \{(a + 2) (a+1)\}}}{2(1)}$

$\sf x = \dfrac{-1\pm \sqrt{(1)^2 - 4 \{(a + 2) (a+1)\}}}{2}$

$\sf x = \dfrac{-1\pm \sqrt{(1)^2 - 4 -(a^2 + 3a+2)\}}}{2(1)}$

$\sf x = \dfrac{-1\pm \sqrt{(1) - 4 -(a^2 + 3a+2)\}}}{2(1)}$

$\sf x = \dfrac{-1\pm\sqrt{1+4a^2+12a+8}}{2}$

$\sf x = \dfrac{-1\pm\sqrt{4a^2+12a+8+1}}{2}$

$\sf x = \dfrac{-1\pm\sqrt{4a^2+12a+9}}{2}$

$\sf x=\dfrac{-1\pm\sqrt{2a}^2 + \sqrt{3}^2}{2}$

$\sf x = \dfrac{-1\pm\sqrt{(2a+3)}^2}{2}$

$\sf x = \dfrac{-1 \pm (2x+3)}2$

$\sf x = \dfrac{2a+2}{2}$

$\sf x = a+1$

or

$\sf x = \dfrac{-2a-4}{2}$

$\sf x = \dfrac{-a-2}{1}$

$\sf x = -(a+2)$

2]

$\sf x =\dfrac{-(-5)\pm\sqrt{(-5)^2 - 4(4)(-3)}}{(2)(4)}$

$\sf x =\dfrac{-(-5)\pm\sqrt{(-5)^2 - 4(4)(-3)}}{8}$

$\sf x =\dfrac{5\pm\sqrt{(-5)^2 - 4(4)(-3)}}{8}$

$\sf x= \dfrac{5\pm\sqrt{25 - 16(-3)}}{8}$

$\sf x = \dfrac{5\pm\sqrt{73}}{8}$

$\sf x = 1.69 \; or \; -0.44$

3]

$\sf\dfrac{x-18}{x} = 6$

$\sf x^2-6x-18=0$

$\sf x = \dfrac{-(-6)\pm\sqrt{-6^2 - 4(1)(-18)}}{(2)(1)}$

$\sf x = \dfrac{-(-6)\pm\sqrt{-6^2 - 4(1)(-18)}}{2}$

$\sf x = \dfrac{-6\pm\sqrt{-6^2 - 4(1)(-18)}}{2}$

$\sf x = \dfrac{6\pm\sqrt{36 - 4(-18)}}{2}$

$\sf x = \dfrac{6 \pm \sqrt{108}}{2}$

$\sf x = \dfrac{6 + 6\sqrt{3}}{2}$

$\sf x = 8.19 \; or \; 2.19$

4]

$\sf x= \dfrac{\sqrt{5}^2 - \sqrt{4 \times 2\times (-5)}}{(2)(2)}$

$\sf x = \dfrac{5 - \sqrt{4 \times 2\times (-5)}}{4}$

$\sf x= \dfrac{5\pm\sqrt{40}}{4}$

$\sf x = \dfrac{5\pm\sqrt{45}}{4}$

$\sf x = 1.11 \; or \; -2.23$