Question

the nature of roots of quadratic equation x2-5x-7=0 are real and equal.​

Answer 1

Answer:

No,the nature of the roots are real and distinct

Step-by-step explanation:

b2-4ac

(-5)2-4×1×(-7)

=25+28

=53

Answer 2

Answer:

The given statement is False.

The nature of roots of quadratic equation $x^{2} -5x-7=0$ are real and distinct.

Step-by-step explanation:

Roots of Quadratic Equation:

Quadratic equations are the equations that contain at least one term with a degree 2. Quadratic equations are in the form $ax^{2} +bx+c=0$ where a, b, and c are some constants.

The root of a quadratic equation is the solution in which the equation is satisfied. That is, it is the value of x that satisfies the equation.

Different methods for finding the root of a quadratic equation:

• Factorization
• Quadratic Formula
• Square Completion
• Graphing

Quadratic Formula:

$x=\frac{-b \pm \sqrt{b^{2}-4ac } }{2a}$

Nature of Roots of Quadratic Equation:

The nature of the root of a quadratic equation is determined using the value of the discriminant,  

$\triangle = b^{2} - 4ac$

• If $\triangle > 0$, roots are real and distinct.
• If $\triangle = 0$, roots are real and equal.
• If $\triangle < 0$, roots are complex and distinct.

Given:

Quadratic equation, $x^{2} - 5x-7=0$

Here, $a=1, \ b=-5, \ c=-7$

Discriminant,

$\triangle = b^{2}-4ac\\\\= (-5)^{2} - (4 \times 1 \times -7)\\\\= 25 - (-28)\\\\= 53 > 0$

Here $\triangle > 0$, i.e., the roots are real and distinct.

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