Question

Solve the following quadratic equations by completing square method :

1. x^2+x-20=0

2. x^2+2x-5=0

Want a well explained answer! ​

Answer 1

Answer :-

→ 1. 4 or -5 .

→ 2. √6 - 1 or -√6 - 1 .

Step-by-step explanation :-

1 . x² + x - 20 = 0 .

$\sf \implies {x}^{2} + x = 20 .\\ \\ \sf \implies {x}^{2} + 2 \times x \times \frac{1}{2} = 20 . \\ \\ \sf \implies {x}^{2} + 2 \times x \times \frac{1}{2} + {( \frac{1}{2} )}^{2} = 20 + {( \frac{1}{2}) }^{2} . \: \: \: \: \{ \tt adding \: {( \frac{1}{2} )}^{2} \: on \: both \: side \} \\ \\ \sf \implies {(x + \frac{1}{2}) }^{2} = 20 + \frac{1}{4} . \\ \\ \sf \implies {(x + \frac{1}{2}) }^{2} = \frac{(80 + 1)}{4} . \\ \\ \sf \implies {(x + \frac{1}{2}) }^{2} = \frac{81}{4} . \\ \\ \sf \implies x + \frac{1}{2} = \pm \sqrt{ \frac{81}{4} } . \\ \\ \sf \implies x + \frac{1}{2} = \pm \frac{9}{2}. \\ \\ \sf \implies x = \frac{ - 1}{2} \pm \frac{9}{2} . \\ \\ \sf \implies x = ( \frac{ - 1}{2} + \frac{9}{2} ) \: \: or \: \: x = ( \frac{ - 1}{2} - \frac{9}{2} ) . \\ \\ \sf \implies x = \frac{8}{2} \: \: or \: \: x = \frac{ - 10}{2} . \\ \\ \huge \pink{ \boxed{ \therefore \it x = 4 \: \: or \: \: x = -5 .}}$

2 . x² + 2x - 5 = 0 .

$\sf \implies {x}^{2} + 2x = 5. \\ \\ \sf \implies {x}^{2} + 2 \times x \times 1 + {1}^{2} = 5 + {1}^{2} . \: \: \: \{ \tt adding \: {1}^{2} \: on \: both \: side \} \\ \\ \sf \implies {(x + 1)}^{2} =6 . \\ \\ \sf \implies x + 1 = \pm \sqrt{6} . \\ \\ \sf \implies x = \pm \sqrt{6} - 1. \\ \\ \huge \orange{ \boxed{ \it \therefore x = \sqrt{6} - 1 \: \: or \: \: - \sqrt{6} - 1.}}$

Hence, it is solved .

Answer 2

Solution :

1.

Solution :

x^2 + x - 20 = 0

=> x^2 + x - 20 = 0

=> x^2 + x = 20

=> x^2 + 2 × x × 1/2 + (1/2)^2 = 20 + (1/2)^2

Now, by adding (1/2)^2 on both the sides.

=> (x + 1/2)^2 = 20 + (1/4)

=> (x + 1/2)^2 = (80 + 1/4)

=> (x + 1/2)^2 = (81/4)

=> √(x + 1/2)^2 = √(81/4)

=> (x + 1/2) = 9/2

=> x + 1/2 = 9/2

=> x = (-1/2 + 9/2) or x = (-1/2 - 9/2)

=> x = 8/2 or x = -10/2

=> x = 4 or x = -5

.°. x = 4, or x = -5.

Answer : x = 4, or x = -5.

2.

Solution :

x^2 + 2x - 5 = 0

=> x^2 + 2x - 5 = 0

=> x^2 + 2x = 5

=> x^2 + 2 × x × 1 + 1^2 = 5 + 1^2

Now, adding 1^2 on both sides.

(x + 1)^2 = 6

=> √(x + 1)^2 = √6

=> x + 1 = √6

=> x = (√6 - 1), or (-√6 - 1)

.°. x = (√6 - 1), or (-√6 - 1)

Answer : x = (√6 - 1), or (-√6 - 1).