Question

Use the quadratic formula to solve the equation x^2-7x-6=0

Answer 1

Answer:

$\frac{7\pm\sqrt{73}} {2}$

Explanation:

We have been given with the quadratic equation $x^2-7x-6=0$

We have general formula to find the roots of a quadratic equation first we find the discriminant with formula

$D=b^{2}-4ac$

and after that to find the variable suppose x we have the formula

$x=\frac{-b\pm\sqrt{D}} {2a}$

And general quadratic equation is

$ax^2+bx+c=0$

On comparing the given quadratic equation with genral quadratic equation we will have values

a=1, b=-7 and c=-6

After substituting these values in the formula we will get  

$D=(-7)^2-4(1)(-6)={73}$

After substituting in the formula to find x we will get

$\frac{-(-7)\pm\sqrt{73}} {2}= \frac{7\pm\sqrt{73}} {2}$

Answer 2

The roots of the equation ${x^2} - 7x - 6 = 0$ are $x = \dfrac{{7 + \sqrt {73} }}{2}{\text{ or   }}x = \dfrac{{7 - \sqrt {73} }}{2}$.

Further explanation:  

The quadratic equation are those equation whose degree is 2.

The general quadratic equation can be expressed as,

$a{x^2} + bx + c = 0$  

Here, $a,b{\text{ and }}c$ are the real numbers.

The roots of the quadratic equation can be found by the quadratic rule.

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

In the above formula ${b^2} - 4ac$ denotes the discriminant.

Therefore, the value of the discriminant cannot be negative because negative value does not exist in the square root for the real numbers.

The negative value in the root is defined only for the complex numbers.

Given:

The given equation is ${x^2} - 7x - 6 = 0$.

Step by step explanation:

Step 1:

The given equation ${x^2} - 7x - 6 = 0$  is the quadratic equation as its degree is 2.

Now to solve the given quadratic equation ${x^2} - 7x - 6 = 0$ by the quadratic rule we need to find the value of the coefficients and constants.

Now compare the given quadratic equation with general quadratic equation to obtain the value of the coefficients and constant as,

$a = 1,b =  - 7,c =  - 6$  

Step 2:

Now substitute the value of $a = 1,b =  - 7,c =  - 6$ in the quadratic formula to obtain the roots of the equation.

$\begin{aligned}x &= \frac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}} \hfill \\x &= \frac{{ - \left( { - 7} \right) \pm \sqrt {{{\left( { - 7} \right)}^2} - 4\left( 1 \right)\left( { - 6} \right)} }}{{2\left( 1 \right)}} \hfill \\x &= \frac{{7 \pm \sqrt {49 + 24} }}{2} \hfill \\ x&= \frac{{7 \pm \sqrt {73} }}{2} \hfill \\\end{aligned}$  

The above equation can be further solved as,

$\begin{aligned}x &= \frac{{7 \pm \sqrt {73} }}{2} \hfill \\x &= \frac{{7 + \sqrt {73} }}{2}{\text{ or   }}x&= \frac{{7 - \sqrt {73} }}{2} \hfill \\\end{aligned}$  

Therefore, the roots of the equation ${x^2} - 7x - 6 = 0$ are $x = \dfrac{{7 + \sqrt {73} }}{2}{\text{ or   }}x = \dfrac{{7 - \sqrt {73} }}{2}$ .

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Answer details:

Grade: Middle school

Subject: Mathematics

Chapter: Quadratic equation

Keywords: linear equation, roots, solution, quadratic equation, coefficients, constants, real number, defined, complex numbers, substitution, general solution, degree, quadratic rule