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.
