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Effectus [21]
3 years ago
8

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

Mathematics
1 answer:
sergij07 [2.7K]3 years ago
8 0

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}

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