Question

what are the solutions of the quadratic equation (x 3)2 = 49? x = –2 and x = –16 x = 2 and x = x = 4 and x = –10 x = 40 and x = –58

Answer 1

Answer:

The solutions of the quadratic equation   x = 4 and x = –10

Step-by-step explanation:

Given quadratic equation $(x+3)^2=49$

We have to find the solution of the given quadratic equation.

Consider the given quadratic equation $(x+3)^2=49$

$\mathrm{For\:}\left(g\left(x\right)\right)^2=f\left(a\right)\mathrm{\:the\:solutions\:are\:}g\left(x\right)=\sqrt{f\left(a\right)},\:\:-\sqrt{f\left(a\right)}$

$x+3=\sqrt{49}$

We have  $\sqrt{49}=\pm 7$

that is $x+3=\pm 7$

Taking both sign separately, we have,

$x+3= 7$ and $x+3=-7$

Simplify , we get,

$x=7-3$ and $x=-7-3$

We get,

$x=4$ and $x=-10$

Thus, The solutions of the quadratic equation   x = 4 and x = –10

Answer 2

The solutions of the quadratic equation ${\left( {x + 3} \right)^2} = 49$ are $\fbox{4}$ and $\fbox{-10}$.

Further explanation:

The general form of the quadratic equation is given by,

$a{x^2} + bx + c$

Here, $a$ is the coefficient of ${x^2}$, $b$ is the coefficient of $x$ and $c$ is the constant term.

Given:

Quadratic equation is ${\left( {x + 3} \right)^2} = 49$.

The options of the solutions of the quadratic equation are $x=-2$,

$x=-16$, $x =2$, $x = 4$, $x =-10$ and $x =-58$.

Calculation:

The solution of the quadratic is the value of the variable at which the value of the polynomial is zero.

A polynomial with degree $2$ is a quadratic equation. The quadratic equation has only two solutions.

The polynomial with degree $n$ has $n$ solution.

Solve the given quadratic equation to obtain the solution of the equation.

$\begin{gathered}{\left( {x + 3} \right)^2} = 49 \\x + 3 = \sqrt {49} \\ \end{gathered}$

Take the square root of $49$ to obtain the value of $x + 3$.

The values of $\sqrt {49}$ are $7$ and $- 7$.

Now solve the equation to obtain the value of $x$.

$\begin{gathered}x + 3 = 7{\text{ and }}x + 3 =- 7 \\ x = 7 - 3{\text{ and}{\text{ }}x =-7-3\ \\ x = 4{\text{}}{\text{ and}}{\text{ }}x =-10\, \\ \end{gathered}$

The values of $x$ are $4$ and $-10$.

Therefore, the solutions of the quadratic equation are ${\left( {x + 3} \right)^2} = 49$ are $4$ and $- 10$.

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Answer details:

Grade: Middle School

Subject: Mathematics

Chapter: Quadratic equation

Keywords: quadratic equation, polynomial, square root, solutions, zeroes, variable, degree, real numbers, coefficients, constant term, general form of quadratic equation.