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While designing the school year book, a teacher asked the student that the length and width of a particular photo is increased by $x$ units each to double the area of the photo. The original photo is 18 cm long and 12 cm wide. Based on the above information, answer the following questions:
(I) Write an algebraic equation depicting the above information.
(II) Write the corresponding quadratic equation in standard form.
(III) What should be the new dimensions of the enlarged photo?
OR
Can any rational value of $x$ make the new area equal to $220\text{cm}^2$
(I) Write an algebraic equation depicting the above information.
(II) Write the corresponding quadratic equation in standard form.
(III) What should be the new dimensions of the enlarged photo?
OR
Can any rational value of $x$ make the new area equal to $220\text{cm}^2$
Show SolutionHide Solution↓
(i) $(18 + x) (12 + x) = 2(18\times12)$
(ii) $x^2 + 30x - 216 = 0$
(iii) Solving : $x^2 + 30x – 216 = 0$
$\Rightarrow (x + 36) (x – 6) = 0$
$x \neq -36 \therefore x = 6$.
new dimensions are $24$ cm $\times 18$ cm
OR
(iii) If $(18+ x) (12 + x) = 220$
then $x^2 + 30x - 4=0$
Here $D = 900 + 16 = 916$ which is not a perfect square.
Thus we can't have any such rational value of $x$.
(ii) $x^2 + 30x - 216 = 0$
(iii) Solving : $x^2 + 30x – 216 = 0$
$\Rightarrow (x + 36) (x – 6) = 0$
$x \neq -36 \therefore x = 6$.
new dimensions are $24$ cm $\times 18$ cm
OR
(iii) If $(18+ x) (12 + x) = 220$
then $x^2 + 30x - 4=0$
Here $D = 900 + 16 = 916$ which is not a perfect square.
Thus we can't have any such rational value of $x$.