Question

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

Answer 1

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 equation\[\sqrt {4 + 2{\text{x - }}{{\text{x}}^2}} = {\text{x}} - 2\]
Hence to solve, square on both side
\[{\left( {\sqrt {4 + 2{\text{x - }}{{\text{x}}^2}} } \right)^2} = {\left( {{\text{x}} - 2} \right)^2}\]
[ Using identity ${\left( {{\text{a - b}}} \right)^2} = \left( {{{\text{a}}^2} + {{\text{b}}^2} - 2{\text{ab}}} \right)$] where a = x and b = 2
\[ \Rightarrow 4 + 2{\text{x - }}{{\text{x}}^2} = {{\text{x}}^2} - 4{\text{x + 4 }}\]
On bringing terms on same side
\[ \Rightarrow 2{{\text{x}}^2} - 6{\text{x + 0 = 0 }}\]
Above expression we can considered as a quadratic equation \[2{{\text{x}}^2} - 6{\text{x + 0 = 0 }}\]
Sridharacharya formula is actually the quadratic formula, used for finding the roots of a quadratic equation \[{\text{a}}{{\text{x}}^2}{\text{ + bx + c = 0}}\] , where a not equal to 0 , & a, b, c are real coefficients of the equation \[{\text{a}}{{\text{x}}^2}{\text{ + bx + c = 0}}\]
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 \[2{{\text{x}}^2} - 6{\text{x + 0 = 0 }}\] with the general quadratic equation \[{\text{a}}{{\text{x}}^2}{\text{ + bx + c = 0}}\] 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 $ \geqslant {\text{0}}$ then it will give real and distinct roots.
Here D ${\text{ = }}\sqrt {{{( - 6)}^2} - 4 \times 2 \times (0)} {\text{ = 6 }} \geqslant {\text{0}}$ 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
\[ \Rightarrow 2{{\text{x}}^2} - 6{\text{x + 0 = 0 }}\]
Taking 2x as a common
\[ \Rightarrow 2{\text{x}}{\text{.(x - 3) = 0 }}\]
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