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Algebraic Expressions And Identities Lessons
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Post a LessonAnswered on 02/02/2024 Learn CBSE - Class 8/Maths/Algebraic Expressions And Identities
Pooja R. Jain
To find the value of x2−15x2−51 at x=−1x=−1, substitute x=−1x=−1 into the expression:
(−1)2−15(−1)2−51
1−151−51
To combine the terms with a common denominator, express 1 as 5555:
55−1555−51
4554
So, the value of x2−15x2−51 at x=−1x=−1 is 4554.
Answered on 02/02/2024 Learn CBSE - Class 8/Maths/Algebraic Expressions And Identities
Pooja R. Jain
To find the value of x2+y2−10x2+y2−10 at x=0x=0 and y=0y=0, substitute these values into the expression:
(0)2+(0)2−10(0)2+(0)2−10
0+0−100+0−10
−10−10
So, the value of x2+y2−10x2+y2−10 at x=0x=0 and y=0y=0 is −10−10.
Answered on 02/02/2024 Learn CBSE - Class 8/Maths/Algebraic Expressions And Identities
Pooja R. Jain
To simplify the expression (a+b+c)(a+b−c)(a+b+c)(a+b−c), you can use the distributive property (FOIL method):
(a+b+c)(a+b−c)(a+b+c)(a+b−c)
=a(a+b−c)+b(a+b−c)+c(a+b−c)=a(a+b−c)+b(a+b−c)+c(a+b−c)
Now, distribute each term:
=a2+ab−ac+ba+b2−bc+ca+cb−c2=a2+ab−ac+ba+b2−bc+ca+cb−c2
Combine like terms:
=a2+2ab−ac+b2−bc−c2=a2+2ab−ac+b2−bc−c2
So, the simplified form of (a+b+c)(a+b−c)(a+b+c)(a+b−c) is a2+2ab−ac+b2−bc−c2a2+2ab−ac+b2−bc−c2.
Answered on 02/02/2024 Learn CBSE - Class 8/Maths/Algebraic Expressions And Identities
Pooja R. Jain
To evaluate (99)2(99)2, simply square the number 99:
(99)2=99×99(99)2=99×99
(99)2=9801(99)2=9801
Therefore, (99)2(99)2 is equal to 9801.
Answered on 02/02/2024 Learn CBSE - Class 8/Maths/Algebraic Expressions And Identities
Pooja R. Jain
To evaluate (95)2(95)2 using identities, you can use the square of a binomial formula:
(a+b)2=a2+2ab+b2(a+b)2=a2+2ab+b2
In this case, a=90a=90 and b=5b=5. Apply the formula:
(95)2=(90+5)2(95)2=(90+5)2
=902+2×90×5+52=902+2×90×5+52
=8100+900+25=8100+900+25
=9025=9025
So, (95)2(95)2 is equal to
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