Solve x2 + 8x − 3 = 0 using the completing-the-square method.
\((a+b)^2=a^2+2ab+b^2\\------------------\\\\x^2+8x-3=0\ \ \ |add\ 3\ to\ both\ sides\\\\x^2+2x\cdot4=3\ \ \ |add\ 4^2\ to\ both\ sides\\\\x^2+2x\cdot4+4^2=3+4^2\\\\(x+4)^2=3+16\\\\(x+4)^2=19\iff x+4=-\sqrt{19}\ or\ x+4=\sqrt{19}\\|subtract\ 4\ from\ both\ sides\\\\\boxed{x=-4-\sqrt{19}\ or\ x=-4+\sqrt{19}}\)
The solutions of the equation \(x^{2}+8x-3=0\) are \(\boxed{0.36\text{ and }-8.36}\).
Further explanation:
Given:
The equation is \(x^{2}+8x-3=0\).
Concept used:
A quadratic function is defined as a function with degree or the highest power of the variable as \(2\).
The standard form of the quadratic equation is,
\(\boxed{ax^{2}+bx+c=0}\) … (1)
The above equation is converted by completing square as follows:
\(\begin{aligned}x^{2}+bx+c+\left(\dfrac{b}{2}\right)^{2}-\left(\dfrac{b}{2}\right)^{2}&=0\\x^{2}+bx+\left(\dfrac{b}{2}\right)^{2}+c-\left(\dfrac{b}{2}\right)^{2}&=0\\ \left(x+\dfrac{b}{2}\right)^{2}+c-\left(\dfrac{b}{2}\right)^{2}&=0\end{aligned}\)
Calculation:
The given equation is \(x^{2}+8x-3=0\).
Now, add and subtract by \(16\) in the left hand side of the equation \(x^{2}+8x-3=0\) and simplify as follows,
\(\begin{aligned}x^{2}+8x-3+16-16&=0\\ (x^{2}+8x+16)-3-16&=0\\ (x+4)^{2}-19&=0\\(x+4)^{2}&=19\end{aligned}\)
Further simplify the above equation as follows,
\(\begin{aligned}(x+4)^{2}&=19\\x+4&=\pm \sqrt{19}\\x+4&=\pm 4.36\\x&=\pm 4.367-4\end{aligned}\)
First take positive sign and obtain the value of \(x\) as follows:
\(\begin{aligned}x&=4.36-4\\&=0.36\end{aligned}\)
Now, take negative sign and obtain the value of \(x\) as follows:
\(\begin{aligned}x&=-4.36-4\\&=-8.36\end{aligned}\)
The values of \(x\) are \(0.36\text{ and }-8.36\).
Therefore, the solutions of the given equation are \(0.36\text{ and }-8.36\).
To check that \(0.36\) is the solution of the equation, substitute \(x=0.36\) in the equation \(x^{2}+8x-3=0\) as,
\(\begin{aligned}(0.36)^{2}+(8\times 0.36)-3\ _{=}^{?}\ 0\\0.1296+2.88-3\ _{=}^{?}\ 0\\0=0\end{aligned}\)
Therefore, \(0.36\) is the solution of the equation \(x^{2}+8x-3=0\).
To check that \(-8.36\) is the solution of the equation, substitute \(x=-8.36\) in the equation \(x^{2}+8x-3=0\) as follows:
\(\begin{aligned}(-8.36)^{2}+(8\times (-8.36))-3\ _{=}^{?}\ 0\\69.8896-66.88-3\ _{=}^{?}\ 0\\0=0\end{aligned}\)
Therefore, \(-8.36\) is the solution of the equation \(x^{2}+8x-3=0\).
Learn more:
1. A problem on quadratic function https://brainly.com/question/1332667
2. A problem on parabola https://brainly.com/question/3213890
Answer details:
Grade: Middle school
Subject: Mathematics
Chapter: Quadratic equation
Keywords: Quadratic equation, complete square, function, equations, degree, highest power, solution, standard form, x2+8x-3, variable.
