Question

Q.3: Find two consecutive positive integers, sum of whose squares is 365.

Q.4: Find the roots of the following quadratic equations, if they exist, by the method of completing the square:

(i) 2x² – 7x +3 = 0

(ii) 2x² + x – 4 = 0

Q.5: The diagonal of a rectangular field is 60 metres more than the shorter side. If the longer side is 30 metres more than the shorter side, find the sides of the field.​

Answer 1

ANSWER(1):

Given:

• Sum of 2 consecutive positive integers = 365

To Find:

• The integers

Solution:

Let the integers be x and (x + 1).

We are given that,

⇒ x² + (x + 1)² = 365

We know that,

⇒ (a + b)² = a² + 2ab + b²

So,

⇒ x² + (x + 1)² = 365

⇒ x² + x² + 2x + 1 = 365

⇒ 2x² + 2x + 1 - 365 = 0

⇒ 2x² + 2x - 364 = 0

⇒ 2(x² + x - 182) = 0

⇒ x² + x - 182 = 0

Splitting the middle term,

⇒ x² + 14x - 13x - 182 = 0

⇒ x(x + 14) - 13(x + 14) = 0

⇒ (x + 14)(x - 13) = 0

So,

⇒ x = -14, 13

As, x is positive. Hence,

⇒ x = 13

And,

⇒ x + 1 = 13 + 1 = 14

Therefore, the integers are 13 and 14.

Formula Used:

• (a + b)² = a² + 2ab + b²

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ANSWER(2):

Given:

• 2x² - 7x + 3 = 0
• 2x² + x - 4 = 0

To Find:

• Roots of each equation using completing the square method.

Solution:

$\text{\bf{1) 2x ^2 - 7x + 3 = 0}}\\\\\text{First, we will make coefficient of x ^2 as 1.}\\\\\text{So, we will divide the equation by 2.}\\\\:\implies\dfrac{2x^2-7x+3}{2}=0\\\\:\implies x^2-\dfrac{7x}{2}+\dfrac{3}{2}=0\\\\:\implies(x)^2-2(x)\left(\dfrac{7}{4}\right)+\dfrac{3}{2}=0\\\\\text{We know that,}\\\\:\hookrightarrow(a-b)^2=a^2-2ab+b^2\\\\\text{Here, a=x and b= \dfrac{7}{4} .}\\\\\text{So, we'll add and subtract \left(\dfrac{7}{4}\right)^2 .}\\\\:\implies(x)^2-2(x)\left(\dfrac{7}{4}\right)+\dfrac{3}{2}+\left(\dfrac{7}{4}\right)^2-\left(\dfrac{7}{4}\right)^2=0\\\\\text{On rearranging,}$

$:\implies(x)^2-2(x)\left(\dfrac{7}{4}\right)+\left(\dfrac{7}{4}\right)^2+\dfrac{3}{2}-\dfrac{49}{16}=0\\\\:\implies\left(x-\dfrac{7}{4}\right)^2+\dfrac{24-49}{16}=0\\\\:\implies\left(x-\dfrac{7}{4}\right)^2-\dfrac{25}{16}=0\\\\:\implies\left(x-\dfrac{7}{4}\right)^2=\dfrac{25}{16}\\\\:\implies\left(x-\dfrac{7}{4}\right)^2=\left(\dfrac{5}{4}\right)^2\\\\\text{Taking square root on both sides,}\\\\:\implies x-\dfrac{7}{4}=\pm\dfrac{5}{4}\\\\:\implies x=\pm\dfrac{5}{4}+\dfrac{7}{4}\\\\:\implies x=\dfrac{5}{4}+\dfrac{7}{4}\:\:\&\:\:x=-\dfrac{5}{4}+\dfrac{7}{4}\\\\:\implies x=\dfrac{12}{4}\:\:\&\:\:x=\dfrac{2}{4}\\\\\bf{:\implies x=3\:\:\&\:\:x=\dfrac{1}{2}}$

$\text{\bf{2) 2x ^2 + x - 4 = 0}}\\\\\text{First, we will make coefficient of x ^2 as 1.}\\\\\text{So, we will divide the equation by 2.}\\\\:\implies\dfrac{2x^2+x-4}{2}=0\\\\:\implies x^2+\dfrac{x}{2}-\dfrac{4}{2}=0\\\\:\implies(x)^2+2(x)\left(\dfrac{1}{4}\right)-2=0\\\\\text{We know that,}\\\\:\hookrightarrow(a+b)^2=a^2+2ab+b^2\\\\\text{Here, a=x and b= \dfrac{1}{4} .}\\\\\text{So, we'll add and subtract \left(\dfrac{1}{4}\right)^2 .}\\\\:\implies(x)^2+2(x)\left(\dfrac{1}{4}\right)-2+\left(\dfrac{1}{4}\right)^2-\left(\dfrac{1}{4}\right)^2=0\\\\\text{On rearranging,}$

$:\implies(x)^2+2(x)\left(\dfrac{1}{4}\right)+\left(\dfrac{1}{4}\right)^2-2-\dfrac{1}{16}=0\\\\:\implies\left(x+\dfrac{1}{4}\right)^2+\dfrac{-32-1}{16}=0\\\\:\implies\left(x+\dfrac{1}{4}\right)^2-\dfrac{33}{16}=0\\\\:\implies\left(x+\dfrac{1}{4}\right)^2=\dfrac{33}{16}\\\\:\implies\left(x+\dfrac{1}{4}\right)^2=\left(\dfrac{\sqrt{33}}{4}\right)^2\\\\\text{Taking square root on both sides,}\\\\:\implies x+\dfrac{1}{4}=\pm\dfrac{\sqrt{33}}{4}\\\\:\implies x=\pm\dfrac{\sqrt{33}}{4}-\dfrac{1}{4}\\\\:\implies x=\dfrac{\sqrt{33}}{4}-\dfrac{1}{4}\:\:\&\:\:x=-\dfrac{\sqrt{33}}{4}-\dfrac{1}{4}\\\\\bf{:\implies x=\dfrac{-\sqrt{33}-1}{4}\:\:\&\:\:x=\dfrac{\sqrt{33}-1}{4}}$

Formulae Used:

• (a + b)² = a² + 2ab + b²
• (a - b)² = a² - 2ab + b²

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ANSWER(3):

Given:

• Diagonal of a rectangle = Shorter side + 60m
• Longer side of rectangle = Shorter side + 30m

To Find:

• Sides of the rectangle

Solution:

Let the length of shorter side be x.

We are given that,

⇒ Diagonal of a rectangle = Shorter side + 60m

⇒ Diagonal = (x + 60)m

And,

⇒ Longer side of rectangle = Shorter side + 30m

⇒ Longer side = (x + 30)m

We know that, the diagonal, length and breadth of a rectangle form a right-angled triangle.

So, by Pythagoras Theorem:

⇒ (Hypotenuse)² = (Base)² + (Height)²

So,

⇒ (Diagonal)² = (Shorter side)² + (Longer side)²

⇒ (x + 60)² = (x)² + (x + 30)²

We know that,

⇒ (a + b)² = a² + 2ab + b²

So,

⇒ (x² + 2(x)(60) + 60²) = (x²) + (x² + 2(x)(30) + 30²)

⇒ x² + 120x + 3600 = x² + x² + 60x + 900

⇒ x² + 120x + 3600 = 2x² + 60x + 900

Transposing LHS to RHS,

⇒ 0 = 2x² + 60x + 900 - (x² + 120x + 3600)

⇒ 2x² + 60x + 900 - x² - 120x - 3600 = 0

⇒ x² - 60x - 2700 = 0

⇒ x² - 90x + 30x - 2700 = 0

⇒ x(x - 90) + 30(x - 90) = 0

⇒ (x - 90)(x + 30) = 0

⇒ x = 90 and (-30)

As, length cannot be negative,

⇒ x = 90

So,

The sides of the rectangle :

• Shorter side = x = 90m
• Bigger side = x + 30 = 90 + 30 = 120m
• Diagonal = x + 60 = 90 + 60 = 150m

Formulae Used:

• (a + b)² = a² + 2ab + b²
• Pythagoras Theorem: (Hypotenuse)² = (Base)² + (Height)²

Answer 2

$\large\color{purple}\underline{\underline{{ \boxed {Solution \: 01} }}}$

Solution:

• Let the two consecutive Numbers be x and x+1.

   Therefore ,

$\tt \: x² + (x+1)² = 365 \\ \tt \: x² + x² +1² + 2 \times x \times 1 = 365\\ \tt \:(because (A+B)² = A² + B²+ 2AB)\\ \tt \: or 2x² + 1 + 2x = 365\\ \tt \: 2x² + 2x = 365 - 1\\ \tt \:2x² + 2x = 364\\ \tt \: 2(x² + x) = 364\\ \tt \:or\: x² + x = 364/2\\ \tt \: or\: x² + x = 182\\ \tt \: or\: x² + x - 182 =0$

     Now Solve the Quadratic Equation ,

$\tt \: x² + 14x - 13x - 182 = 0 \\ \tt \: x (x+14) - 13 (x +14 ) = 0\\ \tt \: ( x- 13 )(x+14)=0$

Therefore , Either x - 13 = 0 or x+14 =0

Since the Consecutive Integers are positive ,

Therefore , x-13 = 0

⇒ x =13

The two consecutive positive Integers are 13 and 14 .

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$\large\color{purple}\underline{\underline{{ \boxed {Solution \: 02} }}}$

$(i) \tt \: 2x {}^{2} – 7x + 3 = 0 \\ \tt \: ⇒ 2x {}^{2} – 7x = - 3$

On dividing both sides of the equation by 2, we get

$\tt \: ⇒ x {}^{2} – 7 \frac{x}{2}  = \frac{ - 3}{2} \\ \tt \: ⇒ x {}^{2} – 2 × x ×   \frac{7}{4} = \frac{ - 3}{2}$

On adding (7/4)2 to both sides of equation, we get

$\tt \: ⇒ (x) {}^{2}- 2 × x × \frac{7}{4} + ( \frac{7}{4} ) {}^{2}  = ( \frac{7}{4} ) {}^{2}  - 3/2 \\ \\ \tt \: ⇒ (x  \frac{ - 7}{4} ) {}^{2}  = \frac{49}{16} - \frac{3}{2} \\ \\ \tt \: ⇒ (x  - \frac{7}{4} ) {}^{2} = \frac{25}{16} \\ \\ \tt \: ⇒ (x  \frac{ - 7}{4} ) = ± \frac{5}{4} \\ \\ \tt \: ⇒ x = \frac{7}{4}  ± \frac{5}{4} \\ \\ \tt \: ⇒ x = \frac{7}{4} + \frac{5}{4} \: or \:  x = \frac{7}{4} - \frac{5}{4} \\ \\ \\ \tt \: ⇒ x = \frac{12}{4} \: or  \: x = \frac{2}{4} \\ \\ \tt \: ⇒ x = 3 \: or \: \frac{1}{2}$

$\tt \: (ii) 2x2 + x – 4 = 0 \\ \tt \: ⇒ 2x2 + x = 4$

On dividing both sides of the equation, we get

$\tt \: ⇒ x2 + x/2 = 2$

On adding (1/4)2 to both sides of the equation, we get

$\\ \tt \: ⇒ (x)2 + 2 × x × 1/4 + (1/4)2 = 2 + (1/4)2\\ \tt \:⇒ (x + 1/4)2 = 33/16\\ \tt \:⇒ x + 1/4 = ± √33/4\\ \tt \:⇒ x = ± √33/4 - 1/4\\ \tt \:⇒ x = ± √33-1/4\\ \tt \:⇒ x = √33-1/4 or x = -√33-1/4$

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$\large\color{purple}\underline{\underline{{ \boxed {Solution \: 03} }}}$

Solution:-

• Let the shorter side of the rectangular field be 'x' meters.
• Longer side will be (x + 30) meters.
• Length of the diagonal will be (x + 60) meters.

The diagonal divides the rectangular into two right angled triangles and the diagonal is the common side of the two triangles and it is also the longest side of the triangles i.e. the hypotenuse.

$\tt \: So, \: by \: Pythagoras \: Theorem, \\ \tt \: (Diagonal)² = (Smaller Side)² + (Longer Side)²\\ \tt \: (x + 60)² = (x)² + (x + 30)²\\ \tt \: x² + 120x + 3600 = x² + x² + 60x + 900\\ \tt \: x² + 60x - 120x + 900 - 3600 = 0\\ \tt \: x² - 60x - 2700 = 0\\ \tt \: x² - 90x + 30x - 2700 = 0\\ \tt \: x(x - 90) + 30(x - 90) = 0\\ \tt \: (x - 90) (x + 30) = 0$

x = 90 because x = - 30 as length cannot be possible.

So the length of the shorter side is 90 meters Length of the longer side is 90 + 30 = 120 meters.

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$\sf \: @WildCat7083$