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Question: Solve for \(x\): \(\dfrac{x}{2} + \dfrac{9}{x} = 4\)...

Solve for xx: x2+9x=4\dfrac{x}{2} + \dfrac{9}{x} = 4

Explanation

Solution

The question is to solve the given equation. We can change the given equation into the quadratic equation by taking L.C.M and interchanging the terms. Let ax2+bx+c=0a{x^2} + bx + c = 0 be a quadratic equation then the roots of this equation are given by −b±b2−4ac2a\dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}. By using the formula −b±b2−4ac2a\dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}} , we can find the roots of the quadratic equation.

Complete answer:
Since given that x2+9x=4\dfrac{x}{2} + \dfrac{9}{x} = 4and then we need to find the value of the unknown variable xx, so we will make use of the basic mathematical operations to simplify further.
Starting with the cross multiplication we have x2+9x=4⇒(x×x)+(9×2)2×x=4\dfrac{x}{2} + \dfrac{9}{x} = 4 \Rightarrow \dfrac{{(x \times x) + (9 \times 2)}}{{2 \times x}} = 4
Using the multiplication operation, we get (x2)+(18)2x=4⇒x2+18=4×2x\dfrac{{({x^2}) + (18)}}{{2x}} = 4 \Rightarrow {x^2} + 18 = 4 \times 2x
Further solving we get x2+18=8x⇒x2−8x+18=0{x^2} + 18 = 8x \Rightarrow {x^2} - 8x + 18 = 0
Here we are asked to solve the given quadratic equation that is we have to find its roots. Since it is an equation of order two it will have two roots. The roots of a quadratic equation can be found by using the formula.
It is given that x2−8x+18=0{x^2} - 8x + 18 = 0 we aim to solve this equation that is we have to find its roots.
We know that the number of roots of an equation is equal to its degree. Here the degree of the given equation is two thus this equation will have two roots.
The roots of a quadratic equation can be found by using the formula −b±b2−4ac2a\dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}} where aa - coefficient of the term x2{x^2}, bb - coefficient of the term xx, and cc - constant term.
First, let us collect the required terms for the formula from the given quadratic equation to solve it.
From the given equation x2−8x+18=0{x^2} - 8x + 18 = 0, we have a=1a = 1, b=−8b = - 8, and c=18c = 18.
On substituting these terms in the formula, we get
x=−(−8)±(−8)2−4(1)(18)2(1)x = \dfrac{{ - \left( { - 8} \right) \pm \sqrt {{{\left( { - 8} \right)}^2} - 4\left( 1 \right)\left( {18} \right)} }}{{2\left( 1 \right)}}
On simplifying this we get
x=8±64−722x = \dfrac{{8 \pm \sqrt {64 - 72} }}{2}
On further simplification we get
x=8±−82x = \dfrac{{8 \pm \sqrt { - 8} }}{2}
=8±2i22= \dfrac{{8 \pm 2i\sqrt 2 }}{2}
⇒x=8+2i22\Rightarrow x = \dfrac{{8 + 2i\sqrt 2 }}{2} and ⇒x=8−2i22 \Rightarrow x = \dfrac{{8 - 2i\sqrt 2 }}{2}
On solving the above, we get
⇒x=4+i2\Rightarrow x = 4 + i\sqrt 2 and ⇒x=4−i2\Rightarrow x = 4 - i\sqrt 2
Thus, we got the roots of the given quadratic equation that is ⇒x=4+i2\Rightarrow x = 4 + i\sqrt 2 and ⇒x=4−i2\Rightarrow x = 4 - i\sqrt 2

Note:
Note that complex imaginary values ii can be reframed in the real form of i2=−1{i^2} = - 1 or in the inverse form as −1=i\sqrt { - 1} = i . The number of roots of an equation depends on its degree. The degree of an equation is the highest power of the unknown variable in that equation. Also, in the quadratic equation, the a=0a = 0 is never possible, because then it will be a linear equation.