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Quadratic Equations - Represent and interpret quadratic equations in standard form ax^2 + bx + c = 0

Grade 10CBSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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A quadratic equation in the variable xx is an equation of the form ax2+bx+c=0ax^2 + bx + c = 0, where a,b,ca, b, c are real numbers and a≠0a \neq 0.

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The standard form of a quadratic equation is ax2+bx+c=0ax^2 + bx + c = 0, where the terms are arranged in descending order of their degrees.

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The degree of a quadratic equation is always 22. If the highest power of xx is not 22 after simplification, it is not a quadratic equation.

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Any equation of the form P(x)=0P(x) = 0, where P(x)P(x) is a polynomial of degree 22, is a quadratic equation.

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A real number α\alpha is said to be a root of the quadratic equation ax2+bx+c=0ax^2 + bx + c = 0 if aα2+bα+c=0a\alpha^2 + b\alpha + c = 0 holds true.

📐Formulae

ax2+bx+c=0, where a≠0ax^2 + bx + c = 0, \text{ where } a \neq 0

D=b2−4acD = b^2 - 4ac

x=−b±b2−4ac2ax = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

💡Examples

Problem 1:

Check whether the following is a quadratic equation: (x−2)2+1=2x−3(x - 2)^2 + 1 = 2x - 3.

Solution:

LHS =(x−2)2+1=x2−4x+4+1=x2−4x+5= (x - 2)^2 + 1 = x^2 - 4x + 4 + 1 = x^2 - 4x + 5 Equating LHS to RHS: x2−4x+5=2x−3x^2 - 4x + 5 = 2x - 3 x2−4x−2x+5+3=0x^2 - 4x - 2x + 5 + 3 = 0 x2−6x+8=0x^2 - 6x + 8 = 0

Explanation:

The simplified equation is in the form ax2+bx+c=0ax^2 + bx + c = 0, where a=1,b=−6,a = 1, b = -6, and c=8c = 8. Since a≠0a \neq 0, it is a quadratic equation.

Problem 2:

Represent the following situation in the form of a quadratic equation: The product of two consecutive positive integers is 306306.

Solution:

Let the first positive integer be xx. Then the next consecutive integer is x+1x + 1. According to the problem: x(x+1)=306x(x + 1) = 306 x2+x=306x^2 + x = 306 x2+x−306=0x^2 + x - 306 = 0

Explanation:

This is the required standard form ax2+bx+c=0ax^2 + bx + c = 0, where a=1,b=1,a=1, b=1, and c=−306c=-306.

Problem 3:

Determine if x=2x = 2 is a root of the equation 3x2−5x−2=03x^2 - 5x - 2 = 0.

Solution:

Substitute x=2x = 2 in the LHS of the equation: LHS =3(2)2−5(2)−2= 3(2)^2 - 5(2) - 2 =3(4)−10−2= 3(4) - 10 - 2 =12−12= 12 - 12 =0= 0

Explanation:

Since LHS == RHS (0=00 = 0), x=2x = 2 satisfies the equation and is therefore a root.