kerryanne78
contestada

Use the quadratic formula to solve the equation.
4x^2 - 10x + 5 = 0
Enter your answers, in simplified radical form.

X=_____ or X=_____​

Respuesta :

r3t40

Note that [tex]+\vee-[/tex] stands for plus or minus.

For the quadratic equation of the form [tex]ax^2+bx+c=0[/tex] the solutions are [tex]x_{1,2}=\dfrac{-b+\vee-\sqrt{b^2-4ac}}{2a}[/tex] that means that for [tex]a=4, b=-10, c=5\Longrightarrow x_{1,2}=\dfrac{-(-10)+\vee-\sqrt{(-10)^2-4\cdot4\cdot5}}{2\cdot4}[/tex] this simplifies to [tex]\boxed{x_1=\dfrac{5+\sqrt{5}}{4}}, \boxed{x_2=\dfrac{5-\sqrt{5}}{4}}[/tex]

Hope this helps.

gmany

Answer:

[tex]\large\boxed{x=\dfrac{5-\sqrt5}{4},\ x=\dfrac{5+\sqrt5}{4}}[/tex]

Step-by-step explanation:

[tex]\text{The quadratic formula for}\ ax^2+bx+c=0\\\\\text{if}\ b^2-4ac<0,\ \text{then the equation has no real solution}\\\\\text{if}\ b^2-4ac=0,\ \text{then the equation has one solution:}\ x=\dfrac{-b}{2a}\\\\\text{if}\ b^2-4ac,\ ,\ \text{then the equation has two solutions:}\ x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}\\\\==========================================[/tex]

[tex]\text{We have the equation:}\ 4x^2-10x+5=0\\\\a=4,\ b=-10,\ c=5\\\\b^2-4ac=(-10)^2-4(4)(5)=100-80=20>0\\\\x=\dfrac{-(-10)\pm\sqrt{20}}{2(4)}=\dfrac{10\pm\sqrt{4\cdot5}}{8}=\dfrac{10\pm\sqrt4\cdot\sqrt5}{8}=\dfrac{10\pm2\sqrt5}{8}\\\\=\dfrac{2(5\pm\sqrt5)}{8}=\dfrac{5\pm\sqrt5}{4}[/tex]

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