ML Aggarwal Solution Class 10 Chapter 5 Quadratic Equations in One Variable MCQs

Q1 Q2 Q3 Q4 Q5 Q6 Q7 Q8 Q9 Q10 Q11 Q12 Q13 Q14 Q15

 MCQs

Question 1

Which of the following is not a quadratic equation ?

(a) (x+2)²=2(x+3)

(b) x²+3x=(–1) (1–3x)

(c) (x+2)(x–1)=x²–2x–3

(d) x³–x²+2x+1=(x+1)3

Sol :

(a) (x + 2)² = 2(x + 3)

⇒ x² + 4x + 4 = 2x + 6

⇒ x² + 4x – 2x + 4 – 6 = 0

⇒ x² + 2x – 2

It is a quadratic equation.

(b) $x^{2}+3 x=(-1)(1-3 x) $

$\Rightarrow x^{2}+3 x=-1+3 x$

$\Rightarrow x^{2}+1=0$

It is also quadratic equation.

(c) $(x+2)(x-1)=x^{2}-2 x-3$

$x^{2}-x+2 x-2=x^{2}-2 x-3$

$x^{2}-x^{2}+x+2 x-2+3=0 \Rightarrow 3 x+1=0$

It is not a quadratic equation.

(d) $x^{3}-x^{2}+2 x+1=(x+1)^{3}$

$=x^{3}+3 x^{2}+3 x+1$

$x^{3}-x^{2}+2 x+1$

$3 x^{2}+x^{2}-2 x-1+3 x+1=0$

$\Rightarrow 4 x^{2}+x=0$

It is a quadratic equation 

Ans : (c)

Question 2

Which of the following is a quadratic equation ?

(a) (x – 2) (x + 1) = (x – 1) (x – 3)

(b) $(x+2)^{3}=2 x\left(x^{2}-1\right)$

(c) $x^{2}+3 x+1=(x-2)^{2}$

(d) $8(x-2)^{3}=(2 x-1)^{3}+3$

Sol :

(a) (x – 2) (x + 1) = (x – 1) (x – 3)
⇒ x2 + x – 2x – 2 = x2 – 3x – x + 3
⇒ 3x + x – 2x + x = 3 + 2
⇒ 3x = 5
It is not a quadratic equation.

(b) $(x+2)^{3}=2 x\left(x^{2}-1\right)$

$x^{3}+6 x^{2}+12 x+8=2 x^{3}-2 x$

$x^{3}+6 x^{2}+12 x+8-2 x^{3}+2 x=0$

$-x^{3}+6 x^{2}+14 x+8=0$

It is not a quadratic equation.

(c) $x^{2}+3 x+1=(x-2)^{2}$

$x^{2}+3 x+1=x^{2}-4 x+4$

⇒3x+1+4 x-4=0 

⇒7x-3=0

It is not a quadratic equation.

(d) $8(x-2)^{3}=(2 x-1)^{3}+3$

$8\left(x^{3}-6 x^{2}+12 x-8\right)$

$=8 x^{3}-12 x^{2}+6 x-1+3$

$8 x^{3}-48 x^{2}+96 x-64-8 x^{3}+12 x^{2}-6 x+1-3=0$

$-36 x^{2}+90 x-66=0$

It is a quadratic equation

Ans : (d)

Question 3

Which of the following equations has 2 as a root ?

(a) $x^{2}-4 x+5=0$

(b) $x^{2}+3 x-12=0$

(c) $2 x^{2}-7 x+6=0$

(d) $3 x^{2}-6 x-2=0$

Sol :

(a) $x^{2}-4 x+5=0$

$\Rightarrow(2)^{2}-4 x^{2}+5=0$

⇒ 4 – 8 + 5 = 0

⇒ 9 – 8 ≠ 0

2 is not its root.

(b) $x^{2}+3 x-12=0$

$ \Rightarrow(2)^{2}-3 \times 2-12=0$

⇒4-6-12=4-18=-14

∴ 2 is not its roots.

(c) $2 x^{2}-7 x+6=0$

$ \Rightarrow 2(2)^{2}-7 \times 2+6=0$

⇒ 8-14+6=0 $

⇒0=0

∴ 2 is its root

(d) $3 x^{2}-6 x-2=0$

$ \Rightarrow 3(2)^{2}-6 \times 2-2=0$

⇒12-12-2=0 

⇒12-14=0

∴ 2 is not its root.

Ans : (c)

Question 4

If $\frac{1}{2}$ is a root of the equation $x^{2}+k x-\frac{5}{4}=0$ then the value of k is

(a) 2

(b) – 2

(c) $\frac{1}{4}$

(d) $\frac{1}{2}$

Sol :

$\frac{1}{2}$ is a root of the equation

$x^{2}+k x-\frac{5}{4}=0$

Substituting the value of $x=\frac{1}{2}$ in the equation 

$\left(\frac{1}{2}\right)^{2}+k \times \frac{1}{2}-\frac{5}{4}=0$

$\Rightarrow \frac{1}{4}+\frac{k}{2}-\frac{5}{4}=0$

$\Rightarrow \frac{k}{2}-1=0$

⇒k=1×2=2

∴k=2

Ans (a)

Question 5

If $\frac{1}{2}$ is a root of the quadratic equation $4 x^{2}-4 k x+k+5=0$ then the value of k is

(a) – 6

(b) – 3

(c) 3

(d) 6

Sol :

$\frac{1}{2}$ is a root of the equation 

$4 x^{2}-4 k x+k+5=0$

Substituting the value of $x=\frac{1}{2}$ in the equation

$4\left(\frac{1}{2}\right)^{2}-4 \times k \times \frac{1}{2}+k+5=0$

1-2k+k+5=0

-k+6=0

k=6

Ans (d)

Question 6

The roots of the equation $x^{2}-3 x-10=0$ are

(a) 2,- 5

(b) – 2, 5

(c) 2, 5

(d) – 2, – 5

Sol :

$x=\frac{-(-3) \pm \sqrt{(-3)^{2}-4 \times 1 \times(-10)}}{2 \times 1}$

$=\frac{3 \pm \sqrt{9+40}}{2}=\frac{3 \pm \sqrt{49}}{2}=\frac{3+7}{2}$

∴$x=\frac{3+7}{2}=5$ and $x=\frac{3-7}{2}=\frac{-4}{2}=-2$

x = 5, – 2 or – 2, 5 

Ans (b)

Question 7

If one root of a quadratic equation with rational coefficients is $\frac{3-\sqrt{5}}{2}$, then the other 

(a) $\frac{-3-\sqrt{5}}{2}$

(b) $\frac{-3+\sqrt{5}}{2}$

(c) $\frac{3+\sqrt{5}}{2}$

(d) $\frac{\sqrt{3}+5}{2}$

Sol :

One root of a quadratic equation is $\frac{3-\sqrt{5}}{2}$ then other root will be $\frac{3+\sqrt{5}}{2}$

Ans (c)

Question 8

If the equation $2 x^{2}-5 x+(k+3)=0$ has equal roots then the value of k is

(a) $\frac{g}{8}$

(b) $-\frac{g}{8}$

(c) $\frac{1}{8}$

(d) $-\frac{1}{8}$

Sol :

$2 x^{2}-5 x+(k+3)=0$

a=2, b=-5, c=k+3

=25-8(k+3)

∴ Roots are equal. 

$\therefore b^{2}-4 a c=0$

∴ 25-8(k+3)=0

⇒25-8k-24=0

⇒1-8k=0 

⇒8 k=1

$\therefore k=\frac{1}{8}$

Ans (c)

Question 9

The value(s) of k for which the quadratic equation $2 x^{2}-k x+k=0$ has equal roots is (are)

(a) 0 only

(b) 4

(c) 8 only

(d) 0, 8

Sol :

$2 x^{2}-k x+k=0$

a=2, b=-k, c=k

$\therefore b^{2}-4 a c=(-k)^{2}-4 \times 2 \times k$

$\quad=k^{2}-8 k$

∴ Roots are equal. $\therefore b^{2}-4 a c=0$

$k^{2}-8 k=0$

⇒k(k-8)=0$

Either k=0

or k-8=0, then k=8

k=0,8

Ans (d)

Question 10

If the equation $3 x^{2}-k x+2 k=0$ roots, then the the value(s) of k is (are)

(a) 6

(b) 0 Only

(c) 24 only

(d) 0

Sol :

$3 x^{2}-k x+2 k=0$

Here, a=3, b=-k, c=2 k

$b^{2}-4 a c=(-k)^{2}-4 \times 3 \times 2 k$

$=k^{2}-24 k$

∴ Roots are equal. $\therefore b^{2}-4 a c=0$

$\therefore k^{2}-24 k=0$

⇒k(k-24)=0

Either k=0 

or k-24=0, then k=24

∴ k=0, 24 

Ans (d)

Question 11

If the equation $\{k+1\} x^{2}-2(k-1) x+1=0$ has equal roots, then the values of k are

(a) 1, 3

(b) 0, 3

(c) 0, 1

(d) 0, 1

Sol :

(k + 1)x² – 2(k – 1)x + 1 = 0

Here, a = k + 1, b = -2(k – 1), c = 1

$\therefore b^{2}-4 a c=[-2(k-1)]^{2}-4(k+1)(1)$

$\quad=4\left(k^{2}-2 k+1\right)-4 k-4$

$\quad=4 k^{2}-8 k+4-4 k-4$

$\quad=4 k^{2}-12 k$

∵ Roots are equal. $\therefore b^{2}-4 a c=0$

$\therefore 4 k^{2}-12 k=0$

⇒4 k(k-3)=0 

⇒ k(k-3)=0

Either k=0 or k-3=0, then k=3

k=0,3(b)

Question 12

If the equation 2x² – 6x + p = 0 has real and different roots, then the values ofp are given by

(a) $p<\frac{9}{2}$

(b)p $\leq \frac{9}{2}$

(c) $p>\frac{9}{2}$

(d) $p \geq \frac{9}{2}$

Sol :

2x² – 6x + p = 0

Here, a = 2, b = -6, c = p

$b^{2}-4 a c=(-6)^{2}-4 \times 2 \times p$

=36-8 p

∵ Roots are real and unequal. 

$\therefore b^{2}-4 a c>0$

⇒ 36-8p>0

⇒ 36-8p>0

⇒ 36>8p

⇒ $\frac{36}{8}>p$

⇒ $p<\frac{36}{8}$

⇒ $p<\frac{9}{2}$

Ans (a)

Question 13

The quadratic equation 2x² – √5x + 1 = 0 has

(a) two distinct real roots

(b) two equal real roots

(c) no real roots

(d) more than two real roots

Sol :

2x² – √5x + 1 = 0

Here, a = 2, b = -√5, c = 1

$b^{2}-4 a c=(-\sqrt{5})^{2}-4 \times 2 \times 1$

=5-8=-3

$\because b^{2}-4 a c<0$

∵ It has no real roots.

Question 14

Which of the following equations has two distinct real roots ?

(a) $2 x^{2}-3 \sqrt{2 x}+\frac{9}{4}=0$

(b) $x^{2}+x-5=0$

(c) $x^{2}+3 x+2 \sqrt{2}=0$

(d) $5 x^{2}-3 x+1=0$

Sol :

(a) $2 x^{2}-3 \sqrt{2} x+\frac{9}{4}=0$

$b^{2}-4 a c=(-3 \sqrt{2})^{2}-4 \times 2 \times \frac{9}{4}=18-18=0$

∵ Roots are real and equal.

(b) $x^{2}+x-5=0$

$b^{2}-4 a c=(1)^{2}-4 \times 1 \times(-5)$

$=1+20=\sqrt{21}>0$

Roots are real and distinct.

Ans (b)

Question 15

Which of the following equations has no real roots ?

(a) x² – 4x + 3√2 = 0

(b) x² + 4x – 3√2 = 0

(c) x² – 4x – 3√2 = 0

(d) 3x² + 4√3x + 4 = 0

Sol :

(a) x² – 4x + 3√2 = 0

b² – 4ac = ( -4)² – 4 × 1 × 3√2

= 16 – 12√2

= 16 – 12(1.4)

= 16 – 16.8

= -0.8

b² – 4ac < 0

Roots are not real.

Ans (a)

SELINA Solution Class 9 Factorisation Chapter 5 Exercise 5E

Question 1

Factorise : ${\displaystyle x^2+\frac{1}{{\left(4x\right)}^2}+1-7x-\frac{7}{2x}}$

Sol:

${\displaystyle x^2+\frac{1}{{\left(4x\right)}^2}+1-7x-\frac{7}{2x}}$

= ${\displaystyle {\left(x\right)}^2+\frac{1}{{\left(2x\right)}^2}+2\times x\times \frac{1}{2x}-7(x+\frac{1}{2x})}$

= ${\displaystyle {(x+\frac{1}{2x})}^2-7(x+\frac{1}{2x})}$

= ${\displaystyle (x+\frac{1}{2x})(x+\frac{1}{2x}-7)}$

= ${\displaystyle (x+\frac{1}{2x})(x-7+\frac{1}{2x})}$

Question 2

Factorise : ${\displaystyle {\left(9a\right)}^2+\frac{1}{{\left(9a\right)}^2}-2-12a+\frac{4}{3a}}$

Sol:

${\displaystyle {\left(9a\right)}^2+\frac{1}{{\left(9a\right)}^2}-2-12a+\frac{4}{3a}}$

= ${\displaystyle {\left(3a\right)}^2+\frac{1}{{\left(3a\right)}^2}-2\times 3a\times \frac{1}{3a}-4(3a-\frac{1}{3a})}$

= ${\displaystyle {(3a-\frac{1}{3a})}^2-4(3a-\frac{1}{3a})}$

= ${\displaystyle (3a-\frac{1}{3a})[(3a-\frac{1}{3a})-4]}$

= ${\displaystyle (3a-\frac{1}{3a})(3a-4-\frac{1}{3a})}$

Question 3

Factorise : ${\displaystyle x^2+\frac{a^2+1}{a}x+1}$

Sol:

${\displaystyle x^2+\frac{a^2+1}{a}x+1=0}$

∴ ${\displaystyle x^2+ax+\frac{1}{a}x+1=0}$

∴ ${\displaystyle x(x+a)+\frac{1}{a}(x+a)=0}$

∴ ${\displaystyle (x+a)(x+\frac{1}{a})=0}$

Question 4

Factorise : ${\displaystyle x^4+y^4-27x^2{y}^2}$

Sol:

${\displaystyle x^4+y^4-27x^2{y}^2}$

= ${\displaystyle {\left(x^2\right)}^2+{\left(y^2\right)}^2-2x^2{y}^2-25x^2{y}^2}$

= ${\displaystyle {(x^2-y^2)}^2-25x^2{y}^2}$

= ${\displaystyle {(x^2-y^2)}^2-{\left(5xy\right)}^2}$      [∵ a2 - b2 = ( a + b )( a - b )]

= ${\displaystyle [(x^2-y^2)+5xy][(x^2-y^2)-5xy]}$

= ${\displaystyle [x^2+5xy-y^2][x^2-5xy-y^2]}$

Question 5

Factorise : 4x⁴ + 9y⁴ + 11x²y²

Sol:

4x⁴ + 9y⁴ + 11x²y²

= (2x²)² + (3y²)² + 12x²y² - x²y²

= (2x² + 3y²)² - x²y²

= (2x² + 3y²)² - (xy)²

= ( 2x² + 3y² - xy )( 2x² + 3y² + xy)          [ ∵ a² - b² = ( a + b )( a - b )]

Question 6

Factorise : ${\displaystyle x^2+\frac{1}{x^2}-3}$

Sol:

${\displaystyle x^2+\frac{1}{x^2}-3}$

= ${\displaystyle x^2+\frac{1}{x^2}-2\times x\times \frac{1}{x}-1}$

= ${\displaystyle {(x-\frac{1}{x})}^2-1}$

= ${\displaystyle {(x-\frac{1}{x})}^2-{\left(1\right)}^2}$

= ${\displaystyle (x-\frac{1}{x}-1)(x-\frac{1}{x}+1)}$   [ ∵ a² - b² = ( a + b )( a - b )]

Question 7

Factorise : a - b - 4a² + 4b²

Sol:

a - b - 4a² + 4b²  
= ( a - b ) - 4( a² - b² )

= ( a - b ) - 4( a - b )( a + b )   [ ∵ a² - b² = ( a + b )( a - b )]

= ( a - b )[ 1 - 4( a + b )]

= ( a - b )[ 1 - 4a - 4b ]

Question 8

Factorise : (2a - 3)² - 2 (2a - 3) (a - 1) + (a - 1)²

Sol:

(2a - 3)² - 2 (2a - 3) (a - 1) + (a - 1)²

= [( 2a - 3 ) - ( a - 1 )]²

= [ 2a - 3 - a + 1 ]²

= ( a - 2 )²

Question 9

Factorise : (a² - 3a) (a² - 3a + 7) + 10

Sol:

(a² - 3a) (a² - 3a + 7) + 10

Let us assume , a² - 3a = x
Then, our polynomial becomes,
( a² - 3a )( a² - 3a + 7 ) + 10
= x( x + 7 ) + 10
= x² + 7x + 10
= x² + 5x + 2x + 10
= x( x + 5 ) + 2 ( x + 5 )
= ( x + 5 )( x + 2 )

By resubstituting the value of x,
= (a² - 3a + 5)( a² - 3a + 2 )

Now, a² - 3a + 5 will have no factor as discriminant is -11 that is less than 0.

And,

∴ a² - 3a + 2 = a² - 2a - a + 2 = a( a - 2) - 1(a - 2) = (a - 1)(a - 2)

So, factor of given polynomial are,

a² - 3a + 2 = a² - 2a - a + 2

= (a² - 3a + 5)(a - 1)(a - 2)

Question 10

Factorise : (a² - a) (4a² - 4a - 5) - 6

Sol:

Let us assume, a² - a = x
Then the given expression is,
(a² - a) (4a² - 4a - 5) - 6
= x( 4x - 5 ) - 6
= 4x² - 5x - 6
= 4x² - 8x + 3x - 6
= 4x( x - 2 ) + 3( x - 2 )
= ( 4x + 3 )( x - 2 )
= [ 4( a² - a ) + 3 ]( a² - a - 2 )          [ resubstitute the value of x ]
= [ 4a² - 4a + 3 ]( a² - a - 2 )
= [ 4a² - 4a + 3 ]( a² - 2a + a - 2 )
= [ 4a² - 4a + 3 ][ a( a - 2 ) + 1( a - 2 )]
= [ 4a² - 4a + 3 ]( a - 2 )( a + 1 )

Question 11

Factorise : x⁴ + y⁴ - 3x²y²

Sol:

x⁴ + y⁴ - 3x²y² 
= x⁴ + y⁴ - 2x²y² - x²y²

= (x²)² + (y²)² - 2x²y² - x²y²

= ( x² - y² )² - (xy)²

= ( x² - y² - xy )( x² - y² + xy )         [ ∵ a² - b² = ( a + b )( a - b )]

Question 12

Factorise : 5a² - b² - 4ab + 7a - 7b

Sol:

5a² - b² - 4ab + 7a - 7b
= 4a² + a² - b² - 4ab + 7a - 7b
= a² - b² + 4a² - 4ab + 7a - 7b
= ( a² - b² ) + 4a( a - b ) + 7( a - b )
= ( a - b )( a + b ) + 4a( a - b ) + 7( a - b )     [ ∵ a² - b² = ( a + b )( a - b ) ]
= ( a - b )[ ( a + b ) + 4a + 7 ]
= ( a - b )[ ( a + b ) + 4a + 7 ]
= ( a - b )[ 5a + b + 7 ]

Question 13

Factorise : 12(3x - 2y)² - 3x + 2y - 1

Sol:

12(3x - 2y)² - 3x + 2y - 1 = 12( 3x - 2y )² - ( 3x - 2y ) - 1
Let us assume that 3x - 2y = a
Then the given expression is
12(3x - 2y)² - 3x + 2y - 1
= 12a² - 3a - 1
= 12a² - 4a + 3a - 1
= 4a( 3a - 1 ) + 1( 3a - 1 )
= ( 4a + 1 )( 3a - 1 )
= [ 4( 3x - 2y ) + 1 ][ 3( 3x - 2y ) - 1 ]    [ resubstitute the value of a ]
= ( 12x - 8y + 1 )( 9x - 6y - 1)

Question 14

Factorise : 4(2x - 3y)² - 8x+12y - 3

Sol:

 4(2x - 3y)² - 8x+12y - 3
=  4(2x - 3y)² - 4(2x - 3y) - 3
Let us assume that 2x - 3y = a
Then the given expression is
 4(2x - 3y)² - 8x+12y - 3 
= 4a² - 4a - 3
= 4a² - 6a + 2a - 3 
= 2a( 2a - 3 ) + 1( 2a - 3)
= ( 2a - 3 )( 2a + 1 )
= [ 2( 2x - 3y ) - 3 ][ 2( 2x - 3y ) + 1 ]
= ( 4x - 6y - 3 )( 4x - 6y + 1 )

Question 15

Factorise : 3 - 5x + 5y - 12(x - y)²

Sol:

3 - 5x + 5y - 12(x - y)²  = 3 - 5( x - y ) - 12(x - y)² 
Let us assume that x - y = a
Then the given expression is
3 - 5x + 5y - 12(x - y)² 
= 3 - 5a - 12a²
= 3 - 9a + 4a - 12a²
= 3( 1 - 3a ) + 4a( 1 - 3a )
= ( 3 + 4a )( 1 - 3a )                 [ resubstitute the value of a ]
= [ 3 + 4( x - y )][ 1 - 3( x - y )]
= ( 3 + 4x - 4y )( 1 - 3x +  3y )

Question 16

Factorise : 9x ² + 3x - 8y - 64y²

Sol:

9x ² + 3x - 8y - 64y² 
= 9x² - 64y² + 3x - 8y
= [ (3x)² - (8y)² ] + ( 3x - 8y )
= [( 3x + 8y )( 3x - 8y )] + ( 3x - 8y )
= ( 3x - 8y )( 3x + 8y + 1 )

Question 17

Factorise : 2√3x² + x - 5√3

Sol:

 2√3x² + x - 5√3

= 2√3x² + 6x - 5x - 5√3

= 2√3x( x + √3 ) - 5( x + √3 )

= ( 2√3x - 5 )( x + √3 )

Question 18

Factorise : ${\displaystyle \frac{1}{4}{(a+b)}^2-\frac{9}{16}{(2a-b)}^2}$ 

Sol:

${\displaystyle \frac{1}{4}{(a+b)}^2-\frac{9}{16}{(2a-b)}^2}$ 

=${\displaystyle \frac{1}{4}[{(a+b)}^2-\frac{9}{4}{(2a-b)}^2]}$

=${\displaystyle \frac{1}{4}[{(a+b)}^2-\left[\frac{3}{2}{(2a-b)}^2\right]]}$

=${\displaystyle \frac{1}{4}\left[(a+b+\frac{3}{2}(2a-b))(a+b-\frac{3}{2}(2a-b))\right]}$

=${\displaystyle \frac{1}{4}\left[(a+b+3a-\frac{3b}{2})(a+b-3a+\frac{3b}{2})\right]}$

= ${\displaystyle \frac{1}{4}\left[(4a-\frac{b}{2})(\frac{5b}{2}-2a)\right]}$

= ${\displaystyle \frac{1}{4}\left[\left(\frac{8a-b}{2}\right)\left(\frac{5b-4a}{2}\right)\right]}$

= ${\displaystyle \frac{1}{4}\left[\frac{1}{4}(8a-b)(5b-4a)\right]}$

= ${\displaystyle \frac{1}{16}(8a-b)(5b-4a)}$

Question 19

Factorise : 2(ab + cd) - a² - b² + c² + d²

Sol:

2(ab + cd) - a² - b² + c² + d² 
= 2ab + 2cd - a² - b² + c² + d² 
= c² + d² + 2cd - a² - b² + 2ab
= ( c² + d² + 2cd ) - ( a² + b² - 2ab )
= ( c + d )² - ( a - b )²
= ( c + d + a - b )( c + d - a + b )

Question 20.1

Find the value of : ( 987 )² - (13)²

Sol:

( 987 )² - (13)² 
= ( 987 + 13 )( 987 - 13)
= 1000 x 974
= 974000

Question 20.2

Find the value of : ( 67.8 )² - ( 32.2 )²

Sol:

( 67.8 )² - ( 32.2 )² 
= ( 67.8 + 32.2 )( 67.8 - 32.2 )
= 100 x 35.6
= 3560

Question 20.3

Find the value of : ${\displaystyle \frac{{\left(6.7\right)}^2-{\left(3.3\right)}^2}{6.7-3.3}}$

Sol:

${\displaystyle \frac{{\left(6.7\right)}^2-{\left(3.3\right)}^2}{6.7-3.3}}$

= ${\displaystyle \frac{(6.7+3.3)(6.7-3.3)}{(6.7-3.3)}}$

= 10

Question 20.4

Find the value of : ${\displaystyle \frac{{\left(18.5\right)}^2-{\left(6.5\right)}^2}{18.5+6.5}}$

Sol:

${\displaystyle \frac{{\left(18.5\right)}^2-{\left(6.5\right)}^2}{18.5+6.5}}$

= ${\displaystyle \frac{(18.5+6.5)(18.5-6.5)}{(18.5+6.5)}}$

= 12