Question

1. Find the zeroes of the polynomial
(i) x2 - 2x - 3 (ii) 49x2 - 64​

Answer 1

Correct Question:

Find the zeroes of the polynomial

(i) $x^2 - 2x - 3$

(ii) $49x^2 - 64$

Final Answer:

(i) The zeroes of the polynomial  $x^2 - 2x - 3$ are 3, and -1.

(ii) The zeroes of the polynomial $49x^2 - 64$ are $\frac{8}{7}$ and $-\frac{8}{7}$.

Given:

The following two polynomials are provided.

(i) $x^2 - 2x - 3$

(ii) $49x^2 - 64$

To Find:

The zeroes of the provided polynomials are to be determined.

Explanation:

The concepts that are important to arrive at the solution to the present problem are as follows

• The term zeroes indicate the roots of the equation.

• The zeroes of the polynomials are the roots of the respective equations when the polynomials are equated to zero.
• A polynomial has the variables with non-negative integral indices.

Step 1 of 3

From the statement available in the problem, write the following equations.

(i) $x^2 - 2x - 3=0$

(ii) $49x^2 - 64=0$

Step 2 of 3

Solve the first equation in the following way.

$x^2 - 2x - 3=0\\x^2 - 3x+x - 3=0\\x(x-3)+1(x-3)=0\\(x-3)(x+1)=0\\x=3,-1$

Step 3 of 3

Solve the second equation in the following way.

$49x^2 - 64=0\\(7x)^2-8^2=0\\(7x+8)(7x-8)=0\\(x+\frac{8}{7})(x-\frac{8}{7})=0\\x=-\frac{8}{7} ,\frac{8}{7}$

Therefore, the required zeroes of the two polynomials are (3,-1) and $(\frac{8}{7},-\frac{8}{7})$ respectively.

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