Question

find the value of k for quadratic equation X square + (2 k + 4) close X + 9 is equal to zero so that they have two equal roots​

Answer 1

Answer:

Hey mate here is ur answer↓

Step-by-step explanation:

Equation:-

x²+(2k+4)x+9=0

D=b²-4ac=0 [°•°Roots are Equal and Real]

(2k+4)²-4×1×9=0

4k²+16k+16-36=0

4k²+16k-10=0 --->÷by 2

2k²+8k-5=0

2k²+8k=5

2k(k+4)=5

k(k+4)=5/2

=>k=5/2

and k=-4

Hope it helps™✓

Answer 2

Question :--- find the value of k for quadratic equation X² + (2 k + 4)X + 9 =0, so that they have two equal roots ?

Concept used :----

If A•x^2 + B•x + C = 0 ,is any quadratic equation,

then its discriminant is given by;

D = B^2 - 4•A•C

• If D = 0 , then the given quadratic equation has real and equal roots.

• If D > 0 , then the given quadratic equation has real and distinct roots.

• If D < 0 , then the given quadratic equation has unreal (imaginary) roots...

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Solution :---

From given Equation , X² + (2 k + 4)X + 9 =0, we have ,

→ A = 1

→ B = (2k+4)

→ C = 9

Above Told concept , if Roots are Equal and real, than ,

→ D = B^2 - 4•A•C = 0

Putting values now , we get,

→ (2k+4)² - 4 * 1 * 9 = 0

using (a+b)² = a² +b² + 2ab now,

→ 4k²+16k+16 - 36 = 0

→ 4k² + 16k - 20 = 0

→ 4(k² + 4k - 5) = 0

→ k² + 4k - 5 = 0

Splitting the Middle Term now,

→ k² + 5k - k - 5 = 0

→ k(k+5) -1(k+5) = 0

→ (k+5)(k-1) = 0

Putting both Equal to zero now,

if k + 5 = 0

→ k = (-5)

if (k-1) = 0

→ k = 1

Hence, value of k will be Either (-5) or 1 , than roots of given Equation have Real and Equal Roots ...