Question

Find the discriminant of the quadratic equation (p+3)x^2 - (5 - p)x + 1 = 0 and hence
determine the value of p for which the roots are real and distinct.
​

Answer 1

EXPLANATION.

Quadratic equation.

⇒ (p + 3)x² - (5 - p)x + 1 = 0.

As we know that,

D = Discriminant Or b² - 4ac.

For real and distinct D = 0.

⇒ [-(5 - p)²] - 4(p + 3)(1) = 0.

⇒ (5 - p)² - 4(p + 3) = 0.

As we know that,

Formula of :

⇒ (x - y)² = x² + y² - 2xy.

Using this formula in equation, we get.

⇒ 25 + p² - 10p - 4p - 12 = 0.

⇒ p² - 14p + 13 = 0.

Factorizes the equation into middle term splits, we get.

⇒ p² - 13p - p + 13 = 0.

⇒ p(p - 13) - 1(p - 13) = 0.

⇒ (p - 1)(p - 13) = 0.

⇒ p = 1 and p = 13.

                                                                                                                       

MORE INFORMATION.

Nature of the factors of the quadratic equation.

(1) = Real and different, if b² - 4ac > 0.

(2) = Rational and different, if b² - 4ac is a perfect square.

(3) = Real and equal, if b² - 4ac = 0.

(4) = If D < 0 Roots are imaginary and unequal or complex conjugate.

Answer 2

Answer:

Given :-

• The quadratic equation is (p + 3)x² - (5 - p)x + 1 = 0

To Find :-

• What is the discriminate of the quadratic equation.
• What is the value of p.

Solution :-

Given equation :

➦ (p + 3)x² - (5 - p)x + 1 = 0

where,

• a = (p + 3)
• b = - (5 - p)
• c = 1

Now, as we know that :

➲ Discriminate (D) = b² - 4ac = 0

Now, by putting the value we get,

↦ - (5 - p)² - 4(p + 3)(1) = 0

↦ - (5 - p)² - 4 × p + 3 × 1 = 0

↦ - (5 - p)² - 4(p + 3) × 1 = 0

➠ - (5 - p)² - 4(p + 3) = 0

Now, by using the formula of (x - y)² we get :

➲ (x - y)² = x² + y² - 2xy

↦ 25 + p² - 10p - 4p - 12 = 0

↦ p² - 10p - 4p - 12 + 25 = 0

↦ p² - 14p - 12 + 25 = 0

↦ p² - 14p + 13 = 0

↦ p² - (13 + 1)p + 13 = 0

↦ p² - 13p - p + 13 = 0

↦ p(p - 13) - 1(p - 13) = 0

↦ (p - 13)(p - 1) = 0

↦ (p - 13) = 0

↦ p - 13 = 0

➠ p = 13

Either,

↦ (p - 1) = 0

↦ p - 1 = 0

➠ p = 1

∴ The value of p is 13 or 1 and the discriminate of the quadratic equation is real and equal.

________________________

EXTRA INFORMATION :-

☛ The general form of equation is ax² + bx + c = 0 then the equation becomes to a linear equation.

☛ The equation in the form of ax² + bx + c = 0, where a , b , c are real numbers and a ≠ 0 is called a quadratic equation in one variables.

☛ b² = 4ac is the discriminate of the equation. Then,

◇ When b² - 4ac = 0 then the roots are real and equal.

◇ When b² - 4ac > 0 then the roots are imaginary and unequal.

◇ When b² - 4ac < 0 then there will be no real roots.