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Question: How do you solve \(2{x^2} - x + 8 = 0\) ?...

How do you solve 2x2x+8=02{x^2} - x + 8 = 0 ?

Explanation

Solution

First compare the given quadratic equation to standard quadratic equation and find the value of numbers aa, bb and cc in given equation. Then, substitute the values of aa, bb and cc in the formula of discriminant and find the discriminant of the given equation. Finally, put the values of aa, bb and DD in the roots of the quadratic equation formula and get the desired result.

Formula used:
The quantity D=b24acD = {b^2} - 4ac is known as the discriminant of the equation ax2+bx+c=0a{x^2} + bx + c = 0 and its roots are given by
x=b±D2ax = \dfrac{{ - b \pm \sqrt D }}{{2a}} or x=b±b24ac2ax = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}

Complete step by step answer:
We know that an equation of the form ax2+bx+c=0a{x^2} + bx + c = 0, a,b,c,xRa,b,c,x \in R, is called a Real Quadratic Equation.
The numbers aa, bb and cc are called the coefficients of the equation.
The quantity D=b24acD = {b^2} - 4ac is known as the discriminant of the equation ax2+bx+c=0a{x^2} + bx + c = 0 and its roots are given by
x=b±D2ax = \dfrac{{ - b \pm \sqrt D }}{{2a}} or x=b±b24ac2ax = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}
So, first compare 2x2x+8=02{x^2} - x + 8 = 0 quadratic equation to standard quadratic equation and find the value of numbers aa, bb and cc.
Comparing 2x2x+8=02{x^2} - x + 8 = 0 with ax2+bx+c=0a{x^2} + bx + c = 0, we get
a=2a = 2, b=1b = - 1 and c=8c = 8
Now, substitute the values of aa, bb and cc in D=b24acD = {b^2} - 4ac and find the discriminant of the given equation.
D=(1)24(2)(8)D = {\left( { - 1} \right)^2} - 4\left( 2 \right)\left( 8 \right)
After simplifying the result, we get
D=164\Rightarrow D = 1 - 64
D=63\Rightarrow D = - 63
Which means the given equation has no real roots.
Now putting the values of aa, bb and DD in x=b±D2ax = \dfrac{{ - b \pm \sqrt D }}{{2a}}, we get
x=(1)±63i2×2x = \dfrac{{ - \left( { - 1} \right) \pm \sqrt {63} i}}{{2 \times 2}}
It can be written as
x=1±63i4\Rightarrow x = \dfrac{{1 \pm \sqrt {63} i}}{4}
x=14+634i\Rightarrow x = \dfrac{1}{4} + \dfrac{{\sqrt {63} }}{4}i and x=14634ix = \dfrac{1}{4} - \dfrac{{\sqrt {63} }}{4}i
So, x=14+634ix = \dfrac{1}{4} + \dfrac{{\sqrt {63} }}{4}i and x=14634ix = \dfrac{1}{4} - \dfrac{{\sqrt {63} }}{4}i are roots/solutions of equation 2x2x+8=02{x^2} - x + 8 = 0.

Therefore, the solutions to the quadratic equation 2x2x+8=02{x^2} - x + 8 = 0 are x=14+634ix = \dfrac{1}{4} + \dfrac{{\sqrt {63} }}{4}i and x=14634ix = \dfrac{1}{4} - \dfrac{{\sqrt {63} }}{4}i..

Note: We can check whether x=14+634ix = \dfrac{1}{4} + \dfrac{{\sqrt {63} }}{4}i and x=14634ix = \dfrac{1}{4} - \dfrac{{\sqrt {63} }}{4}i are roots/solutions of equation 2x2x+8=02{x^2} - x + 8 = 0 by putting the value of xx in given equation.
Putting x=14+634ix = \dfrac{1}{4} + \dfrac{{\sqrt {63} }}{4}i in LHS of equation 2x2x+8=02{x^2} - x + 8 = 0.
LHS=2(14+634i)2(14+634i)+8{\text{LHS}} = 2{\left( {\dfrac{1}{4} + \dfrac{{\sqrt {63} }}{4}i} \right)^2} - \left( {\dfrac{1}{4} + \dfrac{{\sqrt {63} }}{4}i} \right) + 8
On simplification, we get
LHS=2(1166316+638i)14634i+8\Rightarrow {\text{LHS}} = 2\left( {\dfrac{1}{{16}} - \dfrac{{63}}{{16}} + \dfrac{{\sqrt {63} }}{8}i} \right) - \dfrac{1}{4} - \dfrac{{\sqrt {63} }}{4}i + 8
LHS=314+634i+314634i\Rightarrow {\text{LHS}} = - \dfrac{{31}}{4} + \dfrac{{\sqrt {63} }}{4}i + \dfrac{{31}}{4} - \dfrac{{\sqrt {63} }}{4}i
LHS=0\Rightarrow {\text{LHS}} = 0
LHS=RHS\therefore {\text{LHS}} = {\text{RHS}}
Thus, x=14+634ix = \dfrac{1}{4} + \dfrac{{\sqrt {63} }}{4}i is a solution of equation 2x2x+8=02{x^2} - x + 8 = 0.
Putting x=14634ix = \dfrac{1}{4} - \dfrac{{\sqrt {63} }}{4}i in LHS of equation 2x2x+8=02{x^2} - x + 8 = 0.
LHS=2(14634i)2(14634i)+8{\text{LHS}} = 2{\left( {\dfrac{1}{4} - \dfrac{{\sqrt {63} }}{4}i} \right)^2} - \left( {\dfrac{1}{4} - \dfrac{{\sqrt {63} }}{4}i} \right) + 8
On simplification, we get
LHS=2(1166316638i)14+634i+8\Rightarrow {\text{LHS}} = 2\left( {\dfrac{1}{{16}} - \dfrac{{63}}{{16}} - \dfrac{{\sqrt {63} }}{8}i} \right) - \dfrac{1}{4} + \dfrac{{\sqrt {63} }}{4}i + 8
LHS=314634i+314+634i\Rightarrow {\text{LHS}} = - \dfrac{{31}}{4} - \dfrac{{\sqrt {63} }}{4}i + \dfrac{{31}}{4} + \dfrac{{\sqrt {63} }}{4}i
LHS=0\Rightarrow {\text{LHS}} = 0
LHS=RHS\therefore {\text{LHS}} = {\text{RHS}}
Thus, x=14634ix = \dfrac{1}{4} - \dfrac{{\sqrt {63} }}{4}i is a solution of equation 2x2x+8=02{x^2} - x + 8 = 0.
Final solution: Therefore, the solutions to the quadratic equation 2x2x+8=02{x^2} - x + 8 = 0 are x=14+634ix = \dfrac{1}{4} + \dfrac{{\sqrt {63} }}{4}i and x=14634ix = \dfrac{1}{4} - \dfrac{{\sqrt {63} }}{4}i.