**Algebraic expression-**An algebraic expression is an expression made up of variables and constants along with mathematical operators.

**Linear equation in one variable –**When an equation has only one variable of degree one.

Examples: – 3x-9 = 0 and – 2t = 5.

**Linear equation in two variable –**When an equation has two variables both of degree one.

Examples: – 7x+y=8 and – 6p-4q+12=0**.**

- When an equation is represented graphically, it is a straight line that may or may not cut the coordinate axes.

- Every point on the line is a solution for the linear equation and Every solution of the linear equation is a point on the line.

- Certain linear equations exist such that their solution is (0,0). Such equations when represented graphically pass through the origin.

- Linear equations of the form y=a, when represented graphically are lines parallel to the x-axis and a is the y-coordinate of the points in that line and if x=a, when represented graphically are lines parallel to the y-axis and a is the x-coordinate of the points in that line.

y – 2*x* = 1 *x* + 2y = 12

y = 2*x* + 1.

substitute for this variable (y) in the second equation to get a linear equation in one variable, *x* + 2 × (2*x* + 1) = 12⇒ 5 *x* + 2 = 12⇒ *x* = 2

Substitute this value in one of the equations to get the value of the other variable.

y = 2 × 2 + 1⇒y = 5

So, (2,5) is the required solution of the pair of linear equations y – 2*x* = 1 and *x* + 2y = 12.

*x* + 2y = 8 and 2*x* – 3y = 2

**Step 1:**

Make the coefficients of any variable the same by multiplying the equations with constants. Multiplying the first equation by 2, we get,

2*x* + 4y = 16

**Step 2:**

Add or subtract the equations to eliminate one variable, giving a single variable equation. Subtract second equation from the previous equation

2x + 4y = 16

2x – 3y = 2

– + –

———————–

0(x) + 7y =14

**Step 3:**

Solve for one variable and substitute this in any equation to get the other variable.

y = 2,

x = 8 – 2 y⇒ x = 8 – 4⇒ x = 4

(4, 2) is the solution.

a_{1}*x *+ b_{1}*y *+ c_{1}=0 a_{2}x + b_{2}y + c_{2}=0,

x and y can be calculated as:

x = (b_{1}c_{2}-b_{2}c_{1})/(a_{1}b_{2}-a_{2}b_{1})

y = (c_{1}a_{2}-c_{2}a_{1})/(a_{1}b_{2}-a_{2}b_{1})

Some equations may be in a form which can be reduced to a linear equation through substitution.

2/x+3/y=4 5/x-4/y=9

In this case, we may make the substitution

1/x = u and 1/y = v

The pair of equations reduces to

2u + 3v = 4 5u – 4v = 9

The above pair of equations may be solved. After solving, back substitute to get the values of x and y.

A polynomial is an algebraic expression in which the exponent on any variable is a whole number.

5x^{3}+3x+1 is an example of a polynomial. It is an algebraic expression as well.

2x+3√x is an algebraic expression, but not a polynomial. – since the exponent on x is 1/2 which is not a whole number.

For a polynomial in one variable – the highest exponent on the variable in a polynomial is the degree of the polynomial. Example:x^{2}+2x+3 is 2, as the highest power of x in the given expression is x2.

A zero of a polynomial **p(x)** is the value of x for which the value of p(x) is 0. If k is a zero of p(x), then **p(k)=0.** Example: p(x)=x^{2}-3x+2.

When x=1, the value of p(x) will be equal to

p(1)=12-3×1+2

=1 – 3+2 =0

Since p(x)=0 at x=1, we say that 1 is a zero of the polynomial x^{2}– 3x+2

Quadratic polynomials can be factorized by splitting the middle term.

For example, consider the polynomial 2x^{2}−5x+3

The middle term in the polynomial 2x^{2}−5x+3 is -5x. This must be expressed as a sum of two terms such that the product of their coefficients is equal to the product of 2 and 3 (coefficient of x^{2} and the constant term)

-5 can be expressed as (-2)+(-3), as -2×-3=6=2×3

Thus, 2x^{2}-5x+3=2x^{2}-2x-3x+3

Now, identify the common factors in individual groups

2x^{2}-2x-3x+3=2x(x-1)-3(x-1)

Taking (x-1) as the common factor, this can be expressed as:

2x(x-1)-3(x-1)=(x-1)(2x-3)

**For Quadratic Polynomial:**

If α and β are the roots of a quadratic polynomial ax^{2}+bx+c, then,

α + β = -b/a (Sum of zeroes = -coefficient of x /coefficient of x^{2)}

αβ = c/a (Product of zeroes = constant term / coefficient of x^{2)}

**For Cubic Polynomial**

If α,β and γ are the roots of a cubic polynomial ax^{3}+bx^{2}+cx+d, then

α+β+γ = -b/a αβ +βγ +γα = c/a αβγ = -d/a

Step 1: arrange the terms of the dividend and the divisor in the decreasing order of their degrees.

Step 2: To obtain the first term of the quotient, divide the highest degree term of the dividend by the highest degree term of the divisor Then carry out the division process.

Step 3: The remainder from the previous division becomes the dividend for the next step. Repeat this process until the degree of the remainder is less than the degree of the divisor.

1. (a+b)^{2}=a^{2}+2ab+b^{2}

2. (a-b)^{2}=a^{2}-2ab+b^{2}

3. (x+a)(x+b)=x^{2}+(a+b)x+ab

4. a^{2}-b^{2}=(a+b)(a-b)

5. a^{3}-b^{3}=(a-b)(a^{2}+ab+b^{2})

6. a^{3}+b^{3}=(a+b)(a^{2}-ab+b^{2})

7. (a+b)^{3}=a^{3}+3a^{2}b+3ab^{2}+b^{3}

8. (a-b)^{3}=a^{3}-3a^{2}b+3ab^{2}-b^{3}

The polynomial equation whose highest degree is two is called a quadratic equation. It is expressed in the form of:**ax² + bx + c = 0** where x is the unknown variable and a, b and c are the constant terms.

Solving the quadratic equation by factorization method-

Consider a quadratic equation 2×2-5x+3=0

2×2-5x+3=0 ⇒ 2×2-2x-3x+3=0

This step is splitting the middle term

We split the middle term by finding two numbers (-2 and -3) such that their sum is equal to the coefficient of x and their product is equal to the product of the coefficient of x2 and the constant.

(-2) + (-3) = (-5)And (-2) × (-3) = 6

2×2-2x-3x+3=0 => 2x(x-1)-3(x-1)=0 => (x-1)(2x-3)=0

In this step, we have expressed the quadratic polynomial as a product of its factors.

Thus, x = 1 and x =3/2 are the roots of the given quadratic equation.

Consider the quadratic equation 2×2-8x=10**(i)** Express the quadratic equation in standard form.2×2-8x-10=0**(ii)** Divide the equation by the coefficient of x2 to make the coefficient of x2 equal to 1.

x2-4x-5=0**(iii)** Add the square of half of the coefficient of x to both sides of the equation to get an expression of the form x2±2kx+k2. (x2-4x+4)-5=0+4**(iv)** Isolate the above expression, (x±k)2 on the LHS to obtain an equation of the form

(x±k)2=p2 (x-2)2=9**(v)** Take the positive and negative square roots.

X – 2=±3 => x= – 1 or x=5

Quadratic Formula is used to directly obtain the roots of a quadratic equation from the standard form of the equation.For the quadratic equation ax2+bx+c=0,**x= [-b± √(b2-4ac)]/2a**

By substituting the values of a,b and c, we can directly get the roots of the equation.

To find out the standard form of a quadratic equation when the roots are given:

Let α and β be the roots of the quadratic equation ax2+bx+c=0. Then,(x−α)(x−β)=0 On expanding, we get,

x2−(α+β)x+αβ=0, which is the standard form of the quadratic equation.

Here, a=1,b=−(α+β) and c=αβ.

Let α and β be the roots of the quadratic equation ax2+bx+c=0. Then,

Sum of roots =α+β=-b/a

Product of roots =αβ= c/a

**Discriminant- **For a quadratic equation of the form ax2+bx+c=0, the expression b2−4ac is called the **discriminant,** (denoted by **D**), of the quadratic equation.

The **discriminant **determines the **nature of roots **of the quadratic equation based on the **coefficients **of the quadratic equation.

**Nature of roots-**Based on the value of the discriminant, D=b2−4ac, the roots of a quadratic equation can be of three types.

Case 1: If **D>0**, the equation has two** distinct real roots**.

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