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 :

kudzordzifrancis

ANSWER

[tex]x = \frac{ 5 - \sqrt{ 5} }{4} \: or \: \: x = \frac{ 5 + \sqrt{ 5} }{4} [/tex]

EXPLANATION

The given quadratic equation is

[tex]4 {x}^{2} - 10x + 5 = 0[/tex]

We compare this to

[tex]a {x}^{2} + bx + c = 0[/tex]

to get a=4, b=-10, and c=5.

The quadratic formula is given by

[tex]x = \frac{ - b \pm \sqrt{ {b}^{2} - 4ac} }{2a} [/tex]

We substitute these values into the formula to get:

[tex]x = \frac{ - - 10 \pm \sqrt{ {( - 10)}^{2} - 4(4)(5)} }{2(4)} [/tex]

This implies that

[tex]x = \frac{ 10 \pm \sqrt{ 100 - 80} }{8} [/tex]

[tex]x = \frac{ 10 \pm \sqrt{ 20} }{8} [/tex]

[tex]x = \frac{ 10 \pm2 \sqrt{ 5} }{8} [/tex]

[tex]x = \frac{ 5 \pm \sqrt{ 5} }{4} [/tex]

The solutions are:

[tex]x = \frac{ 5 - \sqrt{ 5} }{4} \: or \: \: x = \frac{ 5 + \sqrt{ 5} }{4} [/tex]

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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