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n simplest radical form, what are the solutions to the quadratic equation 6 = x2 – 10x? Quadratic formula: x = x = 5 x = 5 x = 5 x = 5

Respuesta :

butthe
To solve the quadratic equation given, we have two ways either to solve it via quadratic equation or by completing the square. For this case, we just solve it by completing the square.

6 = x² -10x
6 + 25 = x² - 10x +25
31 = (x - 5)²
√31 = √(x - 5)²
√31 = x-5
√31 + 5 = x
x = √31 + 5
SociometricStar

Answer:

[tex]5\pm\sqrt{31}[/tex]

Step-by-step explanation:

The quadratic equation is [tex]6=x^2-10x[/tex]

Subtract 6 to both sides

[tex]x^2-10x-6=0[/tex]

[tex]\mathrm{For\:a\:quadratic\:equation\:of\:the\:form\:}ax^2+bx+c=0\mathrm{\:the\:solutions\:are\:}\\\\x_{1,\:2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}[/tex]

Applying this formula

[tex]\mathrm{For\:}\quad a=1,\:b=-10,\:c=-6:\quad \\\\x_{1,\:2}=\frac{-\left(-10\right)\pm \sqrt{\left(-10\right)^2-4\cdot \:1\left(-6\right)}}{2\cdot \:1}[/tex]

[tex]x=\frac{-\left(-10\right)\pm\sqrt{\left(-10\right)^2-4\cdot \:1\cdot \left(-6\right)}}{2\cdot \:1}\\\\=\frac{10\pm\sqrt{124}}{2\cdot \:1}\\\\=\frac{10\pm2\sqrt{31}}{2}\\=5\pm\sqrt{31}[/tex]

Therefore, the solution of the given quadratic equation is

[tex]5\pm\sqrt{31}[/tex]

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