Question

If a- b =0.9 and ab = 0.36 find a+ b and a^2-b^2

Answer 1

Given:- a - b = 0.9 and ab = 0.36.

To find:- find the value of a+ b and a^2-b^2.

Solution:-

(a - b)=0.9

ab=0.36

(a+b)^2

=(a-b)^2+4ab

=(0.9)^2+4*(0.36) [where, a-b=0.9,ab=0.36]

=2.25

(a+b)^2=2.25

or, (a+b)=√2.25

or, (a+b)= ± 1.5

(a^2-b^2)

= (a+b)(a-b)

= (±1.5) * (0.9) [where, a-b=0.9,a+b=±1.5]

= ± 1.35

Hence, the value of (a+b) is ± 1.5 and (a^2-b^2) is ±1.35.

Answer 2

Step-by-step explanation:

Given that: If

$a - b = 0.9 \\ \\ ab = 0.36$

To find:

$ab \: and \: {a}^{2} - {b}^{2}$

Solution: First of all, square both side

${(a - b)}^{2} = 0.9 \times 0.9 \\ \\ {a}^{2} + {b}^{2} - 2ab = 0.81 \\ \\ now \: to \: convert \: LHS \: to \: ( {a + b)}^{2} \\ \\ add \: and \: subtract \: 2ab \: in \: LHS \\ \\ {a}^{2} + {b}^{2} - 2ab - 2ab + 2ab= 0.81 \\ \\ {a}^{2} + {b}^{2} + 2ab - 4ab = 0.81 \\ \\ {(a + b)}^{2} - 4ab = 0.81 \\ \\ {(a + b)}^{2} = 4ab + 0.81 \\ \\ {(a + b)}^{2} = 4(0.36) + 0.81 \\ \\ {(a + b)}^{2} = 2.25 \\ \\ take \: square \: root \: both \: sides \: \\ (a + b) = ± 1.5$

Now,to find

${a}^{2} - {b}^{2} \\ \\ remember \: the \: identity \\ \\ (a - b)(a + b) = {a}^{2} - {b}^{2} \\ \\ we \: have \: both \: (a - b) \: and \: (a + b)$

Case 1: (a+b) = +1.5

${a}^{2} - {b}^{2} = (a + b)(a - b) \\ = 1.5 \times 0.9 \\ \\ {a}^{2} - {b}^{2}= 1.35$

Case 2: (a+b) = -1.5

${a}^{2} - {b}^{2} = - 1.5 \times 0.9 \\ = - 1.35$

Hope it helps you.