Question

guys please help me with this question and whoever will give me the answer I will mark them as brainlist but I don't want any useless answer​

Answer 1

Question :

Find the value of m for which the given equation has real and equal roots.

x² + 2(m - 1)x + (m + 5) = 0

Answer :

The required value of m is 4 (or) -1

Step-by-step explanation :

Given :

the quadratic equation is x² + 2(m - 1)x + (m + 5) = 0

To find :

the value of m

Solution :

For the quadratic equation ax² + bx + c = 0 ;

the nature of roots is determined by the value of discriminant which is given by D = b² - 4ac

So, for the given quadratic equation,

• a = 1

• b = 2(m - 1)

• c = m + 5

Finding the discriminant,

D = [2(m - 1)]² - 4(1)(m + 5)

D = 4[m² + 1² - 2(m)(1)] - 4(m + 5)

D = 4[m² + 1 - 2m] - 4m - 20

D = 4m² + 4 - 8m - 4m - 20

D = 4m² - 12m - 16

Since the given equation  has real and equal roots, D = 0

4m² - 12m - 16 = 0

4[m² - 3m - 4] = 0

  m² - 3m - 4 = 0/4

  m² - 3m - 4 = 0

  m² + m - 4m - 4 = 0

  m[m + 1] - 4[m + 1] = 0

   [m + 1] [m - 4] = 0

  ⇒ m + 1 = 0 ; m = -1

  ⇒ m - 4 = 0 ; m = 4

The value of m can be either 4 (or) -1

Answer 2

$\sf{Answer}$

Step by step explanation:-

x² +2(m-1) x + m+5 =0

Given that roots are real &Equal

If Discriminant = 0 Roots are real and equal

Discriminant =b²-4ac

Since,

b²-4ac=0

x² +2(m-1) x +(m+5)=0

a = 1

b = 2(m-1)

c = m+5

b²-4ac = 0

[2(m-1)]²-4(1)(m+5) = 0

4(m-1)² -4(m+5) = 0

4(m²-2m+1) -4m-20=0

4m²-8m+4-4m-20=0

4m²-12m-16=0

Take common 4

4(m²-3m-4)=0

m²-3m-4=0

m² -4m+m-4 =0

m(m-4) +1(m-4) =0

(m-4) (m +1) =0

m -4 =0

m= 4

m+1=0

m = -1

So,Required values of m are 4,-1

Know more:-

If D>0 roots are real & distinct

D<0 roots are complex &conjugate

D>0 D is a perfect square then roots are rational and distinct

D<0 D is not a perfect square then roots are irrational and conjugate to each other

D= 0 roots are rational and equal

Know more in brainly:-

Find the positive value of k for which the equation 2x2 + kx + 2 = 0 has real root

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