Question
If $a - b = 10$ and $ab = 11;$ find $a + b.$
✓
Answer
$a - b =10, ab =11$
We know that:
$(a-b)^2=a^2-2 a b+b^2 $
$\Rightarrow(10)^2=a^2+b^2-2 \times 11 $
$\Rightarrow 100=a^2+b^2-22 $
$\Rightarrow a^2+b^2 $
$=100+22 $
$=122
$
Using $(a+b)^2=a^2+b^2+2 a b$, we get
$(a+b)^2 $
$=122+2(11) $
$=122+22 $
$=144 $
$\Rightarrow(a+b) $
$=\sqrt{144} $
$= \pm 12$
We know that:
$(a-b)^2=a^2-2 a b+b^2 $
$\Rightarrow(10)^2=a^2+b^2-2 \times 11 $
$\Rightarrow 100=a^2+b^2-22 $
$\Rightarrow a^2+b^2 $
$=100+22 $
$=122
$
Using $(a+b)^2=a^2+b^2+2 a b$, we get
$(a+b)^2 $
$=122+2(11) $
$=122+22 $
$=144 $
$\Rightarrow(a+b) $
$=\sqrt{144} $
$= \pm 12$
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 the given figure, $A B \| C D, A B=7 \ cm , B D=25 \ cm$ and $C D=17 \ cm$;find the length of side $BC$.
→Prove that the bisectors of opposite angles of a parallelogram are parallel.
A tap can fill a tank in $12$ hours while another tap can fill the same tank in $x$ hours. Both the taps if opened together can fill the tank in $6$ hours and $40$ minutes. Find the time the second tap will take to fill the tank.
→In an equilateral triangle, prove that the centroid and the circumcentre of the triangle coincide.
Evaluate :$\left(\frac{27}{8}\right)^{\frac{2}{3}}-\left(\frac{1}{4}\right)^{-2}+5^0$
→The dimensions of a field are 15 m $\times$ 12 m. A pit 7.5 m $\times$ 6m $\times$ 1.5 m is dug in one corner of the field and the earth removed from it, is evenly spread over the remaining area of the field. Calculate, by how much the level of the field is raised.
→$D$ and $E$ are points on the sides $A B$ and $A C$ of $\triangle A B C$ such that $D E \| B C$ and divides $\triangle A B C$ into two parts, equal in area. Find $\frac{ BD }{ AB }$.
→Two equal chords $A B$ and $C D$ of a circle with center $O$, intersect each other at point $P$ inside the circle.Prove that: $(i) AP = CP; (ii) BP = DP$
→In the diagram, given below, $\triangle ABC$ is right$-$angled at $B$ and $BD$ is perpendicular to $AC.$Find:$(i) \cos \angle DBC;(ii) \cot \angle DBA$
→Simplify the following:$\left(\frac{64 a^{12}}{27 b^6}\right)^{-\frac{2}{3}}$
→