Question

If equation 9x2 + 6px + 4 = 0 has equal roots, then both roots are equal to​

Answer 1

Answer:

±2/3

Step-by-step explanation:

Given,

equation ( quadratic )  is: 9x² + 6px + 4 = 0

Now, its also states that the roots of the equation are equal so,

d = 0

---> b² - 4ac = 0

---> (6p)² - 4 x 9 x 4 = 0

---> 36p² - 36 x 4 = 0

---> 36p² = 36 x 4

---> p² = 4

---> k = ±2

Case 1: When k = 2,

we get a quadratic eq as:

9x² + 6(2) + 4 = 0

--->9x² + 12x + 4 =  0

--->(3x)² + 2 x 2 x 3x + (2)²= 0  ( Identity: (a+b)²= a² + 2ab + b² )

--->(3x + 2)² = 0

--->3x + 2 = 0

---> x = -2/3

Case 2: When k = -2

we get a quadratic eq as:

9x² + 6(-2) + 4 = 0

--->9x² - 12x + 4 =  0

--->(3x)² - 2  x 2 x - 3 + (2)²= 0 ( Identity: (a - b)²= a² - 2ab + b² )

--->(3x - 2)² = 0

--->3x - 2 = 0

x = 2/3

so the two roots of the quadratic equation are -2/3 and 2/3 or can be written as ±2/3

Hope this helped

Answer 2

Given : Equation 9x² + 6px + 4 = 0 has equal roots,

To Find :  both roots are equal

Solution:

Quadratic equation is of the form ax²+bx+c=0  where a  , b and c are real also  a≠0.

D =  b²-4ac is called discriminant.

D >0 roots are real and distinct

D =0 roots are real and equal

D < 0 roots are imaginary ( not real ) and different

9x² + 6px + 4 = 0

a = 9

b = 6p

c  = 4

D = (6p)² - 4(9)(4)

D = 0

=> 36p² - 144 = 0

=> p² - 4 = 0

=> p = ± 2

9x²  +6(±2)x + 4  = 0

9x²  ±12x + 4  = 0

=> (3x  ±  2)² = 0

Hence x = ± 2/3

Both roots can be either 2/3  or  - 2/3  depending  upon value of p.

Another simpler method:

Roots = α , α

Product of roots = α² = 4/9

=> α = ± 2/3

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