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Question

Question: Solve following algebraic equation \[\sqrt {4 + 2{\text{x - }}{{\text{x}}^2}} = {\text{x}} - 2\]...

Solve following algebraic equation
4+2x - x2=x2\sqrt {4 + 2{\text{x - }}{{\text{x}}^2}} = {\text{x}} - 2

Explanation

Solution

Hint: Proceed the solution of this question, first by squaring on both then we can find the value of x either considering it as a quadratic equation or using Zero Product Property, we can also find the desired values of x.

Complete step-by-step answer:
In this question it is given a algebraic equation4+2x - x2=x2\sqrt {4 + 2{\text{x - }}{{\text{x}}^2}} = {\text{x}} - 2
Hence to solve, square on both side
(4+2x - x2)2=(x2)2{\left( {\sqrt {4 + 2{\text{x - }}{{\text{x}}^2}} } \right)^2} = {\left( {{\text{x}} - 2} \right)^2}
[ Using identity (a - b)2=(a2+b22ab){\left( {{\text{a - b}}} \right)^2} = \left( {{{\text{a}}^2} + {{\text{b}}^2} - 2{\text{ab}}} \right)] where a = x and b = 2
4+2x - x2=x24x + 4 \Rightarrow 4 + 2{\text{x - }}{{\text{x}}^2} = {{\text{x}}^2} - 4{\text{x + 4 }}
On bringing terms on same side
2x26x + 0 = 0 \Rightarrow 2{{\text{x}}^2} - 6{\text{x + 0 = 0 }}
Above expression we can considered as a quadratic equation 2x26x + 0 = 0 2{{\text{x}}^2} - 6{\text{x + 0 = 0 }}
Sridharacharya formula is actually the quadratic formula, used for finding the roots of a quadratic equation ax2 + bx + c = 0{\text{a}}{{\text{x}}^2}{\text{ + bx + c = 0}} , where a not equal to 0 , & a, b, c are real coefficients of the equation ax2 + bx + c = 0{\text{a}}{{\text{x}}^2}{\text{ + bx + c = 0}}
Being quadratic it has 2 roots.
X = \dfrac{{\left( { - {\text{b + }}\sqrt {{{\text{b}}^2} - 4{\text{ac}}} } \right)}}{{2{\text{a}}}}{\text{ & }}\dfrac{{\left( { - {\text{b - }}\sqrt {{{\text{b}}^2} - 4{\text{ac}}} } \right)}}{{2{\text{a}}}}...... (1)

On comparing the given equation 2x26x + 0 = 0 2{{\text{x}}^2} - 6{\text{x + 0 = 0 }} with the general quadratic equation ax2 + bx + c = 0{\text{a}}{{\text{x}}^2}{\text{ + bx + c = 0}} we got values of coefficients a = 2, b = -6, c = 0
On putting the value of coefficients a, b, c in equation (1)
{\text{x = }}\dfrac{{\left( { - ( - 6){\text{ + }}\sqrt {{{( - 6)}^2} - 4 \times (2) \times (0)} } \right)}}{{2 \times 2}}{\text{ & }}\dfrac{{\left( { - ( - 6){\text{ - }}\sqrt {{{( - 6)}^2} - 4 \times (2) \times (0)} } \right)}}{{2 \times 2}}
{\text{x = }}\dfrac{{\left( {{\text{6 + 6}}} \right)}}{4}{\text{ = 3 & }}\dfrac{{\left( {{\text{6 - 6}}} \right)}}{4} = 0
We know that if discriminant D 0 \geqslant {\text{0}} then it will give real and distinct roots.
Here D  = (6)24×2×(0) = 6 0{\text{ = }}\sqrt {{{( - 6)}^2} - 4 \times 2 \times (0)} {\text{ = 6 }} \geqslant {\text{0}} Therefore we got two distinct real roots {{\text{x}}_1}{\text{ = 3 & }}{{\text{x}}_2}{\text{ = 0}}
Hence there will be two values of x i.e. 3 and 0 which will satisfy the above equation.

Note- this type of particular question we can also solve by using quadratic equations because the constant term was zero over there. So after this step we can also solve like
2x26x + 0 = 0 \Rightarrow 2{{\text{x}}^2} - 6{\text{x + 0 = 0 }}
Taking 2x as a common
2x.(x - 3) = 0 \Rightarrow 2{\text{x}}{\text{.(x - 3) = 0 }}
Hence with the help of the "Zero Product Property" says that:
If a × b = 0 then either a = 0 or b = 0 (or both a=0 and b=0)
Hence in the above case, we can directly say either 2x=0 or (x-3) =0
\because 2x=0 ⇒ x=0
\because (x-3) =0 ⇒ x=3