Question

zeros of polynomial 6x²+19x-77=0​

Answer 1

Answer:

6x²+19x-77=0.....(Given quadratic equation)

Comparing the above equation with ax²+bx+c

a=6 , b=19 and c=-77

. b²-4ac=(19) ²-4×6×-77

=361-24×-77

=2209

Now, by formula method,

x=-b±√b²-4ac/2a

=-19±√2209/12

therefore, x= -19+√2209/12 or -19-√2209/12

Answer 2

Answer:

The roots of the given quadratic equation are

$\boxed{\red{\sf\:x\:=\:-\:\frac{11}{2}}}\:\:\:\sf\:or\:\:\:\boxed{\red{\sf\:x\:=\:\frac{7}{3}}}$

Step-by-step-explanation:

The given quadratic equation is

$\sf\:6x^{2}\:+\:19x\:-\:77\:=\:0$.

$\therefore\sf\:6x^{2}\:+\:19x\:-\:77\:=\:0\\\\\implies\sf\:6x^{2}\:+\:33x\:-\:14x\:-\:77\:=\:0\\\\\implies\sf\:3x\:(\:2x\:+\:11\:)\:-\:7\:(\:2x\:+\:11\:)\:=\:0\\\\\implies\sf\:(\:2x\:+\:11\:)\:(\:3x\:-\:7\:)\:=\:0\\\\\implies\sf\:(\:2x\:+\:11\:)\:=\:0\:\:\:or\:\:\:(\:3x\:-\:7\:)\:=\:0\\\\\implies\sf\:2x\:+\:11\:=\:0\:\:\:or\:\:\:3x\:-\:7\:=\:0\\\\\implies\sf\:2x\:=\:-\:11\:\:\:or\:\:\:3x\:=\:7\\\\\implies\boxed{\red{\sf\:x\:=\:-\:\frac{11}{2}}}\:\:\:\sf\:or\:\:\:\boxed{\red{\sf\:x\:=\:\frac{7}{3}}}$

Additional Information:

1. Quadratic Equation :

An equation having a degree '2' is called quadratic equation.

The general form of quadratic equation is

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

Where, a, b, c are real numbers and a ≠ 0.

2. Roots of Quadratic Equation:

The roots means nothing but the value of the variable given in the equation.

3. Methods of solving quadratic equation:

There are mainly three methods to solve or find the roots of the quadratic equation.

A) Factorization method

B) Completing square method

C) Formula method

4. Solution of Quadratic Equation by Factorization:

1. Write the given equation in the form $\sf\:ax^{2}\:+\:bx\:+\:c\:=\:0$

2. Find the two linear factors of the $\sf\:LHS$ of the equation.

3. Equate each of those linear factor to zero.

4. Solve each equation obtained in 3 and write the roots of the given quadratic equation.