Question

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

Show work please.

Answer 1

Answer:

The standard form of quadratic equation $ax^2+bx+c = 0$, then the values of x which are the solutions of the equation are given by:

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

Given the equation: $x^2-7x-6=0$

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

then, the solution of the given equation given by:

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

Substitute the values of a, b and c in above we get;

$x=\frac{-(-7)\pm\sqrt{(-7)^2-4(1)(-6)}}{2(1)}$ or

$x=\frac{7\pm\sqrt{49+24}}{2} =\frac{7\pm\sqrt{73}}{2}$

then,

the solutions of the given equation are;

$x=\frac{7+\sqrt{73}}{2}$ , $\frac{7-\sqrt{73}}{2}$.

​

Answer 2

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

Further explanation:  

The equation with degree2 is called the quadratic equation.

The general quadratic equation can be written as,

$a{x^2} + bx + c = 0$  

In the above formula, $a,b{\text{ and }}c$ are the real numbers.

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

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

Here, ${b^2} - 4ac$ denotes the discriminant.

Since, we know that negative value does not exist in the square root for the real numbers.

Therefore, the value of the discriminant cannot be negative.

The negative value in the root is not defined for the real 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 it’s degree is 2.

First we need to find the value of the coefficients and constants.

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

$a = 1,b =  - 7,c =  - 6$  

Step 2:

Now use the quadratic rule to find the given quadratic equation ${x^2} - 7x - 6 = 0$.

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

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

Further simplify the above equation.

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

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