Question

If the roots of the quadratic equation (p+1)x^{2}-6(p+1)x+3(p+9)=0 are equal, find p and then find the roots of this quadratic equation. ​

Answer 1

$\LARGE{ \underline{\underline{ \sf{Required \: answer:}}}}$

GiveN:

• Quadratic p(x) = (p+1)x² - 6(p+1) x+ 3(p+9)= 0
• Roots are real and equal.

To FinD:

• Value of p?
• Roots of the quad. equation?

Step-wise-Step Explanation:

Compare given equation with the general form of quadratic equation, which is ax² + bx + c:

• a = p + 1
• b = -6(p + 1)
• c = 3(p + 9)

We know that when the roots are real and equal, Discriminate 'D' = 0. ( D = b² - 4ac)

⇒ b² - 4ac = 0

Plugging the given values to find p:

⇒ {-6(p + 1)}² - 4 × 3(p + 9)(p + 1) = 0

⇒ 36(p² + 2p + 1) - 12(p² + 10p + 9) = 0

⇒ 36p² + 72p + 36 - 12p² - 120p - 108 = 0

⇒ 24p² - 48p - 72 = 0

⇒ 24(p² - 2p - 3) = 0

⇒ p² - 2p - 3 = 0

Factorising by middle term factorisation,

⇒ p² - 3p + p - 3 = 0

⇒ p(p - 3) + 1(p - 3) = 0

⇒ (p + 1)(p - 3) = 0

Then, p = -1,3

p cannot be -1 because the quad. eq. will no longer remain a quad. eq. Hence, p = 3.

⇒ 4x² - 24x + 36 = 0

⇒ x² - 6x + 9 = 0

Factorising by middle term factorisation,

⇒ x² - 3x - 3x + 9 = 0

⇒ x(x - 3) - 3(x - 3) = 0

⇒ (x - 3)(x - 3) = 0

Then, x = 3,3

Hence,

• The value of p is -3
• The roots are 3 and 3.

Answer 2

♣ Qᴜᴇꜱᴛɪᴏɴ :

• If the roots of the quadratic equation (p + 1)x² - 6(p + 1)x + 3(p + 9) = 0 are equal, find p and then find the roots of this quadratic equation. ​

★═════════════════★  

♣ ᴀɴꜱᴡᴇʀ :

• Roots are x = 3 , 3

★═════════════════★

♣ ᴄᴀʟᴄᴜʟᴀᴛɪᴏɴꜱ :

We know for equal roots Discriminant 'D' = 0. Also, D = b² - 4ac

So : b² - 4ac = 0

∴ D = b² - 4ac = 0

Here :

• a = p +1
• b = -6(p + 1)
• c = 3(p + 9)

Keeping the values in D = b² - 4ac = 0 :

⇒ [-6(p + 1)]² - 4× (p + 1) × 3(p +9) = 0

Apply exponent rule : (-a)ⁿ = aⁿ, where n is even

⇒ 6²(p + 1)² - 12(p + 1)(p +9) = 0

⇒ 36(p + 1)² - 12(p + 1)(p +9) = 0

We know : (a + b)² = a² + b² + 2ab

⇒ 36(p² + 1² + 2 × p × 1) - 12(p + 1)(p +9) = 0

⇒ 36(p² + 1 + 2p) - 12(p + 1)(p +9) = 0

⇒ (36 × p²) + (36 × 1) + (36 × 2p) - 12(p + 1)(p +9) = 0

⇒ 36p² + 36 + 72p  - 12p² - 120p - 108 = 0

⇒ 36p²- 12p²+ 72p - 120p + 36 - 108 = 0

⇒ 24² + 72p - 120p + 36 - 108 = 0

⇒ 24p² - 48p + 36 - 108 = 0

⇒ 24p² - 48p -72  = 0

Taking 24 as common :

⇒ 24(p² - 2p - 3)  = 0

⇒ 24(p + 1)(p - 3)=0

Using the Zero Factor Principle:

If ab = 0 then a = 0 or b = 0 (or both a=0 and b=0)

p + 1 = 0 or p - 3 = 0

p + 1 = 0

⇒ p + 1 - 1 = 0 - 1

⇒ p = -1

p - 3 = 0

⇒ p - 3 + 3 = 0 + 3

⇒ p = 3

∴ p = -1, 3

Neglecting p ≠ -1

∴ p = 3

Already we have :

• a = p + 1
• b = -6(p + 1)
• c = 3(p + 9)

Substituting the value of p :

a = p + 1    

a = 3 + 1

a = 4

b = -6(p + 1)

b = -6(3 + 1)

b = -6(4)

b = -24

c = 3(p + 9)

c = 3(3 + 9)

c = 3(12)

c = 36

Putting these values in quadratic equation : ax² + bx + c = 0

ax² + bx + c = 0

⇒ 4x² + (-24x) + 36 = 0

⇒ 4x² - 24x + 36 = 0

Let's solve with Quadratic Equation !!

$\mathrm{For\:a\:quadratic\:equation\:of\:the\:form\:}ax^2+bx+c=0\mathrm{\:the\:solutions\:are\:}$

$\sf{\displaystyle x_{1,\:2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}}$

For a = 4, b = -24, c = 36

$\sf{\displaystyle x_{1,\:2}=\frac{-\left(-24\right)\pm \sqrt{\left(-24\right)^2-4\cdot \:4\cdot \:36}}{2\cdot \:4}}$

$\sf{\displaystyle x_{1,\:2}=\frac{-\left(-24\right)\pm \sqrt{0}}{2\cdot \:4}}$

$\sf{x=\dfrac{-\left(-24\right)}{2\cdot \:4}}$

$\sf{x=\dfrac{24}{8}}$

x = 3

∴ Roots are x = 3 , 3