Question

If a and b are two real numbers such that $a^2+b^2=7$ and $a^3+b^3=10$, then

• A. greatest value of |a+b| is 5
• B. greatest value of a+b is 4
• C. greatest value of a+b is 1
• D. least value of |a+b| is 1

Answer 1

The correct option is A greatest value of |a+b| is 5

Given $a^3+b^3=10$

Now, $a^3+b^3={(a+b)}^3-3a^{2}b-3a{b}^2$

$\therefore 10={(a+b)}^3-3ab[a+b]$

$\therefore 10=(a+b)[{(a+b)}^2-3ab]$

$\therefore 10=(a+b)[a^2+2ab+b^2-3ab]$

$\therefore 10=(a+b)[(a^2+b^2)-ab]$

But, $a^2+b^2=7$

$\therefore 10=(a+b)[7-ab]$ (1)

Now, Given $a^2+b^2=7$

$\therefore {(a+b)}^2-2ab=7$

$\therefore 2ab={(a+b)}^2-7$

$\therefore ab=\frac{{(a+b)}^2-7}{2}$

Put this value in equation (1), we get,

$(a+b)[7-\frac{{(a+b)}^2-7}{2}]=10$

$\therefore (a+b)\left[\frac{14-[{(a+b)}^2-7]}{2}\right]=10$

$\therefore (a+b)\{14-[{(a+b)}^2-7]\}=20$

$\therefore (a+b)\{14-{(a+b)}^2+7\}=20$

$\therefore (a+b)\{21-{(a+b)}^2\}=20$

$\therefore 21(a+b)-{(a+b)}^3=20$

$\therefore {(a+b)}^3-21(a+b)+20=0$

Put $(a+b)=x$

$\therefore x^3-21x+20=0$

Put $x=1$ in above equation.

$\therefore LHS=1-21+20=0$

$=RHS$

Thus, $x=1$ is one factor of given equation.

By using synthetic division, we can form quadratic equation as,

$x^2+x-20=0$

$\therefore x^2+5x-4x-20=0$

$\therefore x(x+5)-4(x+5)=0$

$\therefore (x+5)(x-4)=0$

$\therefore x=4$ and $x=-5$

$\therefore |a+b|=1,4,5$

Thus, greatest value of $|a+b|$ is 5