Question

the product of two consecutive integers is 132. frame a quadratic equation and solve to find the integers

Answer 1

S O L U T I O N :

Let the two consecutive integers be r and r+1 respectively.

A/q

$\longrightarrow\rm{r\times (r+1)=132}\\\\\longrightarrow\rm{r^{2} +r=132}\\\\\longrightarrow\rm{r^{2} +r-132=0}$

$\boxed{\bf{By \:using\:quadratic\:formula\::}}}}$

As the get polynomial compared with ax² + bx + c;

• a = 1
• b = 1
• c = -132

Now;

$\longrightarrow\rm{x=\dfrac{-b\pm\sqrt{b^{2} -4ac} }{2a} }\\\\\\\longrightarrow\rm{x=\dfrac{-1\pm\sqrt{(1)^{2} -4\times 1\times (-132)} }{2\times 1} }\\\\\\\longrightarrow\rm{x=\dfrac{-1\pm\sqrt{1+528} }{2} }\\\\\\\longrightarrow\rm{x=\dfrac{-1\pm\sqrt{529} }{2} }\\\\\\\longrightarrow\rm{x=\dfrac{-1\pm23}{2} }\\\\\\\longrightarrow\rm{x=\dfrac{-1+23}{2} \:\:\:Or\:\:\:x=\dfrac{-1-23}{2} }\\\\\\\longrightarrow\rm{x=\cancel{\dfrac{22}{2}} \:\:\;Or\:\:\:x=\cancel{\dfrac{-24}{2} }}$

$\longrightarrow\bf{x=11\:\:\:Or\:\:\:x\neq -12}$

Thus;

$\bullet\:\sf{1st\:integer\:=x=\boxed{11}}}}\\\\\bullet\:\sf{2nd\:integer\:=(x+1)=(11+1)=\boxed{12}}}}$

Answer 2

ɢɪᴠᴇɴ :-

The product of two consecutive integers is 132. Frame a quadratic equation.

ᴛᴏ ғɪɴᴅ :-

• Integers

sᴏʟᴜᴛɪᴏɴ :-

Let one integer be x

then ,

Second integer = (x + 1)

✞ According to Question :-

• Product of two consecutive integers = 132

➱ x(x + 1) = 132

➱ x² + x = 132

➱ x² + x - 132 = 0

➱ x² + 12x - 11x - 132 = 0

➱ x(x + 12) - 11(x + 12) = 0

➱ (x - 11)(x + 12) = 0

➱ (x - 11) = 0. or (x + 12) = 0

➱ x = 11 or x = -12

Here we get two different values of x

So,

Both solutions of x satisfy the conditions of question

Hence,

CASE -1. (If we take x = 11)

• First Integer = x = 11
• Second Integer = (x + 1) = 11 + 1 = 12

CASE - 2 (If we take x = -12)

• First integer = x = -12
• Second Integer = (x + 1) = -12 + 1 = -11