Question

If Zeba were younger by 5 years than what she really is, then the square of her age (in years) would have been 11 more than five times her actual age. What is her age now?

Answer 1

Hint:Proceed the solution of this question by assuming the present age of Zeba, and trying to form a quadratic equation as per the condition given in the following question. And the roots of that quadratic equation will give the actual age of Zeba.

Complete step-by-step answer:
Let x= Zeba’s present actual age
In the question it is given that,
Zeba were younger by 5 years than what she really, can be write in expression as (x - 5) years
Square of her age [i.e. square of (x - 5) ] would have been 11 more than five times her actual age (x)
equation can be formed as
⇒${({\text{x - 5}})^2} = 11{\text{ + 5x}}$ ( Using${\left( {{\text{a - b}}} \right)^2} = {{\text{a}}^2} + {{\text{b}}^2} - 2{\text{ab}}$) here, a = x and b = 5
⇒${{\text{x}}^2} + 25 - 10x = 11{\text{ + 5x}}$
On arranging one side and forming quadratic equation
⇒${{\text{x}}^2} + 25 - 10x - {\text{5x - 11}} = 0$
⇒${{\text{x}}^2} - 15x + 14 = 0$
We have, a quadratic equation ${{\text{x}}^2} - 15x + 14 = 0$
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 ${{\text{x}}^2} - 15x + 14 = 0$ with the general quadratic equation \[{\text{a}}{{\text{x}}^2}{\text{ + bx + c = 0}}\] we got values of coefficients a = 1, b = -15, c = 14
On putting the value of coefficients a, b, c in equation (1)
⇒${\text{x = }}\dfrac{{\left( { - ( - 15){\text{ + }}\sqrt {{{( - 15)}^2} - 4 \times (1) \times (14)} } \right)}}{{2 \times 1}}{\text{ & }}\dfrac{{\left( { - ( - 15){\text{ - }}\sqrt {{{( - 15)}^2} - 4 \times (1) \times (14)} } \right)}}{{2 \times 1}}$
⇒${\text{x = }}\dfrac{{\left( {{\text{15 + }}\sqrt {225 - 56} } \right)}}{2}{\text{ & }}\dfrac{{\left( {{\text{15 - }}\sqrt {225 - 56} } \right)}}{2}$
⇒${\text{x = }}\dfrac{{\left( {{\text{15 + }}\sqrt {169} } \right)}}{2}{\text{ & }}\dfrac{{\left( {{\text{15 - }}\sqrt {169} } \right)}}{2}$
⇒${\text{x = }}\dfrac{{\left( {{\text{15 + 13}}} \right)}}{2}{\text{ & }}\dfrac{{\left( {{\text{15 - 13}}} \right)}}{2}$
⇒${\text{x = }}\dfrac{{28}}{2}{\text{ = 14, & }}\dfrac{{\text{2}}}{2} = 1$

⇒${{\text{x}}_1}{\text{ = }}14{\text{ & }}{{\text{x}}_2}{\text{ = }}1$
Since her age can't be 1, So it is 14.
Zeba’s present actual age=14

Note: In such types of particular questions, we got two roots from the quadratic equation formed by the given condition in the question. Here, sometimes both roots are the correct answer and sometimes only one root we select as a correct answer. It depends on roots whether they are satisfying the given condition given in the question or not. If both roots satisfied the given conditions, then both will be selected as the correct answer. For example- In the above question, for x=1,
When Zeba was younger by 5 years, then her age can be expressed as (x - 5) i.e. (1-5) = - 4, which is a negative quantity and Age can’t be negative, hence x=1 can’t be the right answer.