# Question #7ab87

Apr 30, 2017

$x = 3$

#### Explanation:

Notice that the equation is of the form

${x}^{2} - s x + p$

where $s = a + b$ and $p = a b$, in this case $a = 3 , b = 4$. So we have that the polynomial factorises as

$\left(x - a\right) \left(x - b\right) = 0$

hence the roots are $3$ and $4$, so the lesser root is $x = 3$

If you don't see it straightforward, don't worry: you can proceed with the standard formula for finding roots of quadratic equations:

${x}_{12} = \frac{s \pm \sqrt{{s}^{2} - 4 p}}{2} = \frac{7 \pm \sqrt{49 - 48}}{2} R i g h t a r r o w {x}_{1} = 3 , {x}_{2} = 4$