Question

★ Here are some problems :

1: $\boxed{\sf{ x² - ( k + 4 )x + 2k + 5 \:=\: 0}}$
2: $\boxed{\sf{( k - 12 )x² + 2 ( k - 12 )x + 2 \:=\: 0}}$

*Solve using $\sf{\red{b²\:-\:4ac\:=\:0}}$

ㅤㅤㅤOR

Name some famous books of Lakshminath Bezbarua

[ above question is of Eng. subject ]*​

Answer 1

Step-by-step explanation:

1)

Given equation is x^2-(k+4)x+2k+5 = 0

It can be written as

x^2-(k+4) x +(2k+5)=0

On comparing with the standard quadratic equation ax^2+bx+c = 0

a = 1

b = -(k+4)

c = 2k+5

If the given equation has real and equal roots then its discriminant must be equal to zero.

The discriminant of the quadratic equation ax^2+bx+c = 0 is D = b^2-4ac

=>b^2-4ac=0

=>[-(k+4)]^2-4(1)(2k+5)=0

=>(k^2+8k+16)-(8k+20)=0

=>k^2+8k+16-8k-20=0

=>k^2-4 = 0

=>k^2=4

=>k=±√4

=>k = ±2

The values of k are 2 and -2

________________________________

2)

Given equation is (k-12)x^2+2(k-12)x+2 = 0

On comparing with the standard quadratic equation ax^2+bx+c = 0

a = k -12

b = 2(k-12)

c = 2

If the given equation has real and equal roots then its discriminant must be equal to zero.

The discriminant of the quadratic equation ax^2+bx+c = 0 is D = b^2-4ac

=>b^2-4ac=0

=>[2(k-12)]^2 -4(k-12)(2)=0

=>4(k-12)^2-8(k-12)=0

=>4(k-12)(k-12)-8(k-12) = 0

=>4(k-12)[k-12-2] =0

=>4(k-12)(k-14)=0

=>(k-12)(k-14)=0/4

=>(k-12)(k-14)=0

=>k-12 = 0 or k-14 = 0

=>k = 12 or k =14

If k=12 then the quadratic equation doesn't exist.

k≠12

k=14

The value of k = 14

_______________________________

Using formulae:-

• the standard quadratic equation ax^2+bx+c = 0
• The discriminant of the quadratic equation ax^2+bx+c = 0 is D = b^2-4ac
• If D> 0 it has real and distinct roots
• If D<0 it has no real roots
• If D = 0 it has real and equal roots.
• (a+b)^2 =a^2+2ab+b^2
• A quadratic equation has at most two roots.

(or)

Famous books of Lakshminath Bezbarua

1.Junuka

2.Surabhi

3.Mahapurux Sri Sankardev Aru Sri Madhabdev

4.Baakhar

5.Burhi aair xadhu

Answer 2

$\boxed{\text{Answer - 1}}$

$\huge\bf Given$

a = 1

b = -(k+4)

c = 2k+5

By using ↓

$\sf{\blue{b²\:-\:4ac\:=\:0}}$

$\bf →(-k+4)² — 4 (1) (2k+5) \:=\: 0)$

$\bf → k²+8K+16—8k+20\:=\: 0)$

$\bf →k²— 4 \:=\: 0)$

$\bf →k = ± 2)$

______________________

$\boxed{\text{Answer - 2}}$

$\huge\bf Given$

a = k-12

b = 2 (k-12) = 2k — 24

c = 2

By using ↓

$\sf{\red{b²\:-\:4ac\:=\:0}}$

$\bf→ (2k-24)² — 4(k-12)(2)$

$\bf→ 4k²-96k +576 – 8k -96$

$\bf→ 4k²-104k +480$

$\bf→ k= 20, 6$

~Note - I had solved the "4k²-104k +480" quadratic equation in attached pic…