Exercise 4.2 Unit 4 Algebraic Expressions and Algebraic Formulas

Let’s solve each of the given problems step by step:

(i) If a + b = 10 and a – b = 6, then find the value of (a² + b²).

We’ll start by squaring both equations to get the values of a² and b²:

(a + b)^2 = a² + 2ab + b²
(a – b)^2 = a² – 2ab + b²

Given that a + b = 10 and a – b = 6, we can use these values to solve for a² and b²:

(10)^2 = a² + 2ab + b² [Substitute a + b = 10]
36 = a² + 2ab + b² … (i)

(6)^2 = a² – 2ab + b² [Substitute a – b = 6]
36 = a² – 2ab + b² … (ii)

Now, add equations (i) and (ii) to eliminate the 2ab term:

36 + 36 = 2a² + 2b²
72 = 2(a² + b²)

Finally, divide both sides by 2:

(a² + b²) = 72 / 2
(a² + b²) = 36

So, the value of (a² + b²) is 36.

(ii) If u + b = 5 and u – b = √17, then find the value of ab.

We need to solve for u and b. Let’s add the two given equations to eliminate b:

(u + b) + (u – b) = 5 + √17
2u = 5 + √17

Now, solve for u:

u = (5 + √17) / 2

Next, subtract the second equation from the first to solve for b:

(u + b) – (u – b) = 5 – √17
2b = 5 – √17

Now, solve for b:

b = (5 – √17) / 2

Now, find the value of ab:

ab = u * b
ab = ((5 + √17) / 2) * ((5 – √17) / 2)

Simplify:

ab = (25 – 17) / 4
ab = 8 / 4
ab = 2

So, the value of ab is 2.

(iii) If a² + b² + c² = 45 and a + b + c = -1, then find the value of ab + be + ca.

To find the value of ab + be + ca, we need to square the second equation (a + b + c = -1) and subtract it from the first equation (a² + b² + c² = 45):

(a + b + c)^2 = (-1)^2 = 1

Subtracting from the first equation:

a² + b² + c² – (a + b + c)^2 = 45 – 1
a² + b² + c² – 1 = 44

Now, substitute a + b + c = -1:

a² + b² + c² = 44 + 1
a² + b² + c² = 45

We can see that this equation is the same as the given equation in the question. Therefore, ab + be + ca = 0.

(iv) If m + n + p = 10 and mn + np + mp = 27, then find the value of m² + n² + p².

We can use the identity (m + n + p)^2 = m² + n² + p² + 2(mn + np + mp) to find the value of m² + n² + p².

Given that m + n + p = 10 and mn + np + mp = 27, we have:

(m + n + p)^2 = (10)^2 = 100

Substituting the given value of mn + np + mp:

m² + n² + p² + 2(27) = 100
m² + n² + p² + 54 = 100

Now, subtract 54 from both sides:

m² + n² + p² = 100 – 54
m² + n² + p² = 46

So, the value of m² + n² + p² is 46.

Leave a Comment