Mathematics
Grade6
Easy

Question

Simplify fraction numerator 1 over denominator 2 end fraction x to the power of 2 end exponent plus fraction numerator 1 over denominator 4 end fraction x to the power of 2 end exponent plus 5

  1. 3 x to the power of 2 end exponent    
  2. fraction numerator 3 x to the power of 2 end exponent over denominator 4 end fraction plus 5    
  3. fraction numerator 3 x to the power of 2 end exponent over denominator 4 end fraction    
  4. 5    

Hint:

This is an algebraic expression. An algebraic expression consists of two components: a variable and a constant. A variable is a quantity that doesn’t have any fixed value. It takes value according to the conditions. It is denoted by the alphabet. A constant is a quantity with a fixed value. In this question, we are given an expression. We have perform the required operations and simply the expression.

The correct answer is: fraction numerator 3 x to the power of 2 end exponent over denominator 4 end fraction plus 5


    The given expression is 1 half x squared space plus space 1 fourth x squared space plus space 5
    There is one variable in the above expression. The variable is: ‘x’. There are three terms in the above expression. We have to perform the operations between them and simplify the expression.
    If we see, first and second term contains the same variable raised to same power 2. Such terms are called “like terms.” And the terms with different variables are called “unlike terms.”
    The third term is a constant.
    For like terms, we can perform the operations between their coefficients by keeping the variable constant. It means we can treat each term as a constant and perform operations like normal operations on constants.
    For the above expression, we can perform an operation between the first term and second term as both the terms as they have the same variable ‘x2’. We will keep ‘x2 ' constant and perform addition between the coefficients.
    Let’s see the expression
    1 half x squared space plus space 1 fourth x squared space plus space 5 space equals space 2 over 4 x squared space plus space 1 fourth x squared space plus space 5
    To make the denominators same of the first term we multiplied and divided the first term by 2
    Solving further we get,
    1 half x squared space plus space 1 fourth x squared space plus space 5 space equals space 3 over 4 x squared space plus space 5
    Therefore, the value of the equation is  3 over 4 x squared space plus space 5

    We have to be careful about the powers of the variables too. If the variable is same and power is different, we cannot perform the operation any further.

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