**Q)** 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?

(iv) Can any rational value of x make the new area equal to 220 cm^{2}

**Ans:**

**VIDEO SOLUTION**

**STEP BY STEP SOLUTION**

**(i) Algebraic Expression:**

Original area of Photo A = 18 x 12 cm^{2}

Its given that by increasing each dimension by x units, area get doubled.

**Hence, (18 + x) (12 + x) = 2 x (18 x 12)**

**(ii) Quadratic equation:**

The standard quadratic equation is in the form of a x^{2} + b x + c = 0

We will solve the above expression in the above form.

(18 + x) (12 + x) = 2 x (18 x 12)

⇒ x^{2} + 30 x + 216 = 2 x 216

**⇒ x ^{2} + 30 x – 216 = 0 **

**(iii) Solving the equation:**

By solving the above equation, we get

⇒ x^{2} + 30 x – 216 = 0

⇒ x^{2} + 36 x – 6 x – 216 = 0

⇒ x (x + 36) – 6 (x + 36) = 0

⇒ (x + 36) (x – 6) = 0

⇒ x = – 36, x = 6

Since x ≠ -36, therefore x = 6

∵ Since old photo’s dimensions were 18 cm and 12 cm

and new photo gets increased by x = 6 cm on each side

**∴ New dimensions of photo are 18 + 6 and 12 + 6 i.e. 24 cm & 18 cm**

**(iv) Rational value of X:**

If by increasing each dimension by x units, the area of photo is enlarged to 220 cm^{2}, then

(18 + x) (12 + x) = 220

⇒ x^{2} + 30 x + 216 = 220

⇒ x^{2} + 30 x – 4 = 0

Now to solve above quadratic equation, we need to find root values:

We know that d = b^{2 }– 4ac

= 900 +16 = 916

Since 916 is not a perfect square.

**Therefore we can not have any rational value of x to make the new area 220cm ^{2}.**

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