Question
If $a - b = 4$ and $a + b = 6$; find $(i) a^2+ b^2(ii) ab$
✓
Answer
$(i)$ We know that,
$(a-b)^2=a^2-2 a b+b^2$
Rewrite the above identity as,
$a^2+b^2=(a-b)+2 a b \dots....(1)$
Similarly, we know that, $(a+b)^2=a^2+2 a b+b^2$
Rewrite the above identity as,
$a^2+b^2=(a+b)^2-2 a b\dots .....(2)$
Adding the equations $( 1 )$ and $( 2 ),$ we have
$2\left(a^2+b^2\right)=(a-b)^2+2 a b+(a+b)^2-2 a b$
$ \Rightarrow 2\left(a^2+b^2\right)=(a-b)^2+(a+b)^2$
$\Rightarrow\left(a^2+b^2\right)=\frac{1}{2}\left[(a-b)^2+(a+b)^2\right] \dots....(3)$
Given that $a+b=6 ; a-b=4$
Substitute the values of $(a+b)$ and $(a-b)$ in equation (3), we have
$a^2+b^2=\frac{1}{2}\left[(4)^2+(6)^2\right]$
$ =\frac{1}{2}[16+36]$
$ =\frac{52}{2}$
$ \Rightarrow\left(a^2+b^2\right)=26 \dots.....(4)$
From equation $(4)$, we have
$\mathrm{a} 2+\mathrm{b} 2=26$
Consider the identity,
$(a-b) 2=a 2+b 2-2 a b\dots ....(5)$
Substitute the value $a-b=4$ and $a 2+b 2=26$
in the above equation, we have
$(4) 2=26-2 a b$
$ \Rightarrow 2 a b=26-16$
$ \Rightarrow 2 a b=10$
$ \Rightarrow a b=5$
$(a-b)^2=a^2-2 a b+b^2$
Rewrite the above identity as,
$a^2+b^2=(a-b)+2 a b \dots....(1)$
Similarly, we know that, $(a+b)^2=a^2+2 a b+b^2$
Rewrite the above identity as,
$a^2+b^2=(a+b)^2-2 a b\dots .....(2)$
Adding the equations $( 1 )$ and $( 2 ),$ we have
$2\left(a^2+b^2\right)=(a-b)^2+2 a b+(a+b)^2-2 a b$
$ \Rightarrow 2\left(a^2+b^2\right)=(a-b)^2+(a+b)^2$
$\Rightarrow\left(a^2+b^2\right)=\frac{1}{2}\left[(a-b)^2+(a+b)^2\right] \dots....(3)$
Given that $a+b=6 ; a-b=4$
Substitute the values of $(a+b)$ and $(a-b)$ in equation (3), we have
$a^2+b^2=\frac{1}{2}\left[(4)^2+(6)^2\right]$
$ =\frac{1}{2}[16+36]$
$ =\frac{52}{2}$
$ \Rightarrow\left(a^2+b^2\right)=26 \dots.....(4)$
From equation $(4)$, we have
$\mathrm{a} 2+\mathrm{b} 2=26$
Consider the identity,
$(a-b) 2=a 2+b 2-2 a b\dots ....(5)$
Substitute the value $a-b=4$ and $a 2+b 2=26$
in the above equation, we have
$(4) 2=26-2 a b$
$ \Rightarrow 2 a b=26-16$
$ \Rightarrow 2 a b=10$
$ \Rightarrow a b=5$
Need a full question paper?
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
Explore more
Similar questions
In $\triangle ABC, BE$ and $CF$ are medians. $P$ is a point on $BE$ produced such that $BE = EP$ and $Q$ is a point on $CF$ produced such that $CF = FQ$. Prove that : $A$ is the mid $-$ point of $PQ$.
→Prove that the quadrilateral formed by joining the mid$-$points of consecutive sides of a rectangle is a rhombus.
→A manufacturer sells an article to a dealer at a profit of $10\%$. the dealer sells to the retailer at a profit of $8\%$ and the retailer sells to the customer at a profit of$ 5\%$. If the customer pays $Rs.12474$ for the article, calculate the cost of manufacturing the same.
→Given: $4 \cot A = 3$find :$(i) \sin A;(ii)\sec A;(iii)\operatorname{cosec}^2A - \cot^2A.$
→In the given figure, equal chords AB and CD of a circle with centre O , cut at right angles at P . If L and M are mid-points of $A B$ and $C D$ respectively, prove that OLPM is a square.
→A dealer marks his goods $25\%$ above the cost price and then allows $10\%$ discount on it. What is the cost price of an article on which he gains $Rs. 575$?
→Find the base of an isosceles triangle whose area is $192\ cm^2$ and the length of one of the equal sides is $20\ cm.$
→If $a ^2-3 a -1=0$ and $a \neq 0$, find $: a +\frac{1}{ a }$
→Construct a triangle using the following data: $PQ + PR = 10.6 \ cm, QR = 4.8 \ cm$ and $\angle R = 45^\circ $
→The sum of digit of a two digit number is $11$. If the digit at ten's place is increased by $5$ and the digit at unit place is decreased by $5$, the digits of the number are found to be reversed. Find the original number.
→