# How do you write a quadratic equation in standard form that has two solutions, 10 and -4 with the leading coefficient must be 1?

Oct 6, 2016

The reqd. qudr. eqn. is $f \left(x\right) = {x}^{2} - 6 x - 40 = 0.$

#### Explanation:

Let $f \left(x\right) = 0$ be the desired qudr. eqn.

It has two solns. $10 \text{ & } - 4$.

Therefore, $\left(x - 10\right) \mathmr{and} \left(x - \left(- 4\right)\right) = \left(x + 4\right)$ are the factors of $f \left(x\right)$.

$\Rightarrow f \left(x\right) = k \left(x - 10\right) \left(x + 4\right) , k \in \mathbb{R} - \left\{0\right\} .$

$\Rightarrow \text{ leading co-eff of }$f$\text{ is } k .$

But, given that, $k = 1.$

$f \left(x\right) = \left(x - 10\right) \left(x + 4\right)$

Hence, the reqd. qudr. eqn. is $f \left(x\right) = {x}^{2} - 6 x - 40 = 0.$