Question

Solve using quadratic formula for 2x^(2)-10x+5=0

Answer 1

Answer:

$x=\frac{5+\sqrt{15}}{2},\:x=\frac{5-\sqrt{15}}{2}$

Step-by-step explanation:

$\mathrm{For\:a\:quadratic\:equation\:of\:the\:form\:}ax^2+bx+c=0\mathrm{\:the\:solutions\:are\:}$

$x_{1,\:2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

$\mathrm{For\:}\quad a=2,\:b=-10,\:c=5$

$x_{1,\:2}=\frac{-\left(-10\right)\pm \sqrt{\left(-10\right)^2-4\cdot \:2\cdot \:5}}{2\cdot \:2}$

$\sqrt{\left(-10\right)^2-4\cdot \:2\cdot \:5}$

Apply exponent rule: (-a)^n=a^n, if n is even

$\left(-10\right)^2=10^2$

$=\sqrt{10^2-4\cdot \:2\cdot \:5}$

$\mathrm{Multiply\:the\:numbers:}\:4\cdot \:2\cdot \:5=40$

$=\sqrt{10^2-40}$

$10^2=100$

$=\sqrt{100-40}$

$\mathrm{Subtract\:the\:numbers:}\:100-40=60$

$=\sqrt{60}$

$60\:\mathrm{divides\:by}\:2\quad \:60=30\cdot \:2$

$=2\cdot \:30$

$30\:\mathrm{divides\:by}\:2\quad \:30=15\cdot \:2$

$=2\cdot \:2\cdot \:15$

$15\:\mathrm{divides\:by}\:3\quad \:15=5\cdot \:3$

$=2\cdot \:2\cdot \:3\cdot \:5$

$2,\:3,\:5\mathrm{\:are\:all\:prime\:numbers,\:therefore\:no\:further\:factorization\:is\:possible}$

$=2\cdot \:2\cdot \:3\cdot \:5$

$=2^2\cdot \:3\cdot \:5$

$=\sqrt{2^2\cdot \:3\cdot \:5}$

Apply Radical Rule:

$=\sqrt{2^2}\sqrt{3\cdot \:5}$

Apply Radical Rule:

$\sqrt{2^2}=2$

$=2\sqrt{3\cdot \:5}$

$\mathrm{Refine}$

$=2\sqrt{15}$

$x_{1,\:2}=\frac{-\left(-10\right)\pm \:2\sqrt{15}}{2\cdot \:2}$

$\mathrm{Separate\:the\:solutions}$

$x_1=\frac{-\left(-10\right)+2\sqrt{15}}{2\cdot \:2},\:x_2=\frac{-\left(-10\right)-2\sqrt{15}}{2\cdot \:2}$

$\frac{-\left(-10\right)+2\sqrt{15}}{2\cdot \:2}$

Apply Rule -(-a)=a

$=\frac{10+2\sqrt{15}}{2\cdot \:2}$

$\mathrm{Multiply\:the\:numbers:}\:2\cdot \:2=4$

$=\frac{10+2\sqrt{15}}{4}$

$=\frac{2\left(5+\sqrt{15}\right)}{4}$

$\mathrm{Cancel\:the\:common\:factor:}\:2$

$=\frac{5+\sqrt{15}}{2}$

$\frac{-\left(-10\right)-2\sqrt{15}}{2\cdot \:2}$

$=\frac{10-2\sqrt{15}}{2\cdot \:2}$

$\mathrm{Multiply\:the\:numbers:}\:2\cdot \:2=4$

$=\frac{10-2\sqrt{15}}{4}$

$=\frac{2\left(5-\sqrt{15}\right)}{4}$

$\mathrm{Cancel\:the\:common\:factor:}\:2$

$=\frac{5-\sqrt{15}}{2}$

$\mathrm{The\:solutions\:to\:the\:quadratic\:equation\:are:}$

$x=\frac{5+\sqrt{15}}{2},\:x=\frac{5-\sqrt{15}}{2}$