Question
Question: Solve following algebraic equation \[\sqrt {4 + 2{\text{x - }}{{\text{x}}^2}} = {\text{x}} - 2\]...
Solve following algebraic equation
4+2x - x2=x−2
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=x−2
Hence to solve, square on both side
(4+2x - x2)2=(x−2)2
[ Using identity (a - b)2=(a2+b2−2ab)] where a = x and b = 2
⇒4+2x - x2=x2−4x + 4
On bringing terms on same side
⇒2x2−6x + 0 = 0
Above expression we can considered as a quadratic equation 2x2−6x + 0 = 0
Sridharacharya formula is actually the quadratic formula, used for finding the roots of a quadratic equation ax2 + bx + c = 0 , where a not equal to 0 , & a, b, c are real coefficients of the equation ax2 + 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 2x2−6x + 0 = 0 with the general quadratic equation ax2 + 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 then it will give real and distinct roots.
Here D = (−6)2−4×2×(0) = 6 ⩾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
⇒2x2−6x + 0 = 0
Taking 2x as a common
⇒2x.(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
∵ 2x=0 ⇒ x=0
∵ (x-3) =0 ⇒ x=3