Worksheet on Nature of Roots of Quadratic Equation

We know the nature of the roots of a quadratic equation depends completely on the value of its discriminant. In this article, we will discuss about nature of the roots of quadratic equations and solve some problems.

What is a Quadratic Equation?

A quadratic equation is a second-degree polynomial equation in the form ax² + bx + c = 0, where a, b, and c are constants and a is not equal to 0.

Important Formulas

Various quadratic equations important formulas are:

• Root Formula (x) = {-b± √( b ²-4ac)}/2a
• Discriminant (Δ) = b ²-4ac

Nature of Roots:

• If Δ > 0, there are two distinct real roots.
• If Δ = 0, there is one real root (a repeated root).
• If Δ < 0, there are two complex roots.

Practice Questions Nature of Roots of Quadratic Equations

Problem 1: What can you say about the roots of the equation x²+2x-4=0?

Solution:

Discriminant is D = 2²-4(1)(-4) = 20 > 0

Thus, the equation has real and distinct roots. Let us evaluate them using the quadratic formula:

x = -b±√D/2a = -2±√20/2

= -1±√5

Problem 2:The length of a rectangle is less than twice its breadth by 1 cm. The length of its diagonal is 17 cm. Find its length and breadth.

Solution:

L = 2B – 1, Diagonal = 17. D² = L² + B²

⇒ 289 = (2B –1)² + B²

⇒ B = 8 and L = 15

Also 8, 15, 17 is a Pythagorean triplet.

Problem 3: The sum of the squares of two consecutive natural numbers is 145. Find those numbers.

Solution:

x² + (x + 1)² = 145

⇒2x ² + 2x + 1 = 145

⇒2x ² + 2x - 144 = 0

⇒x ² + x - 72 = 0

⇒x(x + 9) - 8(x + 9) = 0

⇒x = -9, 8

Problem 4: If mx2+9x-1=0 has real roots, find the possible values of m.

Solution:

For real roots, the discriminant of the given equation must be non-negative:

D ≥ 0 ⇒ 9² - 4(m)(-1) ≥ 0

⇒ 81+4m ≥ 0

⇒ m ≤ -81/4

Problem 5: The equation (p+1)x²+2(p+3)x+(p+8)=0 has equal roots. Find the possible values of p.

Solution:

Discriminant of the given equation must be 0

4(p+3)²-4(p+1)(p+8)=0

⇒-3p+1 = 0

⇒ p = 1/3

Problem 6: Find the possible values of t if the roots of the equation x²+t²=8x+6t are real.

Solution:

In standard form, the given equation is

x² -8x + t² - 6t = 0

For the roots to be real, the discriminant should be non-negative:

(-8)² ≥ 4(t²-6t)

⇒ t²-6t ≤ 16

⇒ (t-8)(t+2) ≤ 0

⇒ t≤-2 or t ≥ 8

Problem 7: Find the values which m can take if the roots of (m-3)x²-2mx+5m=0 are real.

Solution:

Discriminant must be non-negative:

D = (-2m)²-4(m-3)(5m) ≥ 0

⇒4m²-4(5m²-15m) ≥ 0

⇒m²-5m²+15m ≥ 0

⇒4m²-15m ≤ 0

⇒4m(m-15/4) ≤ 0

⇒0 ≤ m ≤ 15/4

Problem 8: For what values of k does 5x²+kx+5=0 have real roots?

Solution:

For real roots, the discriminant should be non-negative: D ≥ 0

⇒ k²-4(5)(5) ≥ 0

⇒ k²-100 ≥ 0

⇒ k ≤ -10 or k ≥ 10

Worksheet on Nature of Roots of Quadratic Equation

The worksheet on nature of roots of quadratic equation is added in form of image below:

You can download the worksheet pdf from here-Worksheet: Nature of roots of quadratic equation

Ans 1. Quadratic equation 6x²- 13x + 4 = 0 has two distinct real roots.

x_1 = \frac{{13 + \sqrt{73}}}{12}, x_2 = \frac{{13 - \sqrt{73}}}{12}

Ans 2. Quadratic equation 25x²- 10x + 1 = 0 has one real and repeated root.

x= \frac{1}{5}

Ans 3. Quadratic equation x²+ 2√3 x - 9 = 0 has has two distinct real roots.

x_{1}= \sqrt{3}, x_{2}= \text{-3}\sqrt{3}

Ans 4. The numbers are 4 and 6.

Ans 5. Length: 12 cm and Breadth: 5 cm

Ans 6. The number of trees lengthways is 40 and the number of trees breadthways is 35.

Ans 7. So, the roots of the equation bx ² + ax + 1 = 0 are - \frac{1}{α} \ \ and \ \ \frac{1}{β}

Ans 8. Quadratic equation 7x²- 9x + 2 = 0 has two distinct real roots.

x = 1 and x = \frac{2}{7}

Ans 9. Quadratic equation x²- ax + b2= 0 has

• a) Two distinct real roots if a^{2}\gt \text{4b}^{2}
• b) One real root (repeated) if a^{2}= \text{4b}^{2}
• c) Two complex conjugate roots if a^{2}\lt \text{4b}^{2}

Ans 10. Quadratic equation 2x²+ 8x + 9 = 0 has complex and non-real roots.

x_1 = \frac{{-2 + \sqrt{2}i}}{2}, \ \ \ \ \ x_2 = \frac{{-2 - \sqrt{2}i}}{2}