Question

Write the sum of real roots of the equation : x2 + 51x-6 = 0



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Answer 1

given :-

x² + 51x - 6

here, we have to find the sum of it's real roots.

we can either find it's roots and then we can sum up them. or we can directly do one thing.

finding roots is a long method.

lemme tell u another method.

sum of roots = -b/a

standard form of quadratic polynomial is = ax² + bx + c

here a = 1, b = 51 and c = -6

hence, sum of it's roots = -b/a

= -51/1

= -51

Answer 2

Solution:

We have to find:

Sum of real roots of the given equation:

$\implies$ $\sf{x^{2}+51x-6}$

We know that:

$\boxed{\sf{Sum\:of\:roots = \frac{ - b}{a}}}$

Here:

a = 1, b = 51 and c = -6

Hence:

The sum of its roots:

$\implies$ $\frac{ - b}{a}$

$\implies$ $\frac{ - 51}{1}$

$\implies$ $- 51$

Therefore:

Final answer: -51

Extra information:

What is a Quadratic equation ?

$\bullet$ Quadratic equation is a polynomial, whose highest power is the square of a variable $\sf{(x^{2}, y^{2}\:etc.)}$

Standard form of quadratic equation:

$\implies$ $\boxed{\sf{{ax}^{2} + bx + c = 0}}$