# How do you solve \frac { 1} { 4} x ^ { 2} = 3x - \frac { 25} { 2}?

Jun 11, 2018

no solutions $\phi$

#### Explanation:

$\setminus \frac{1}{4} {x}^{2} = 3 x - \setminus \frac{25}{2}$

multiply both sides by 4 to remove the fractions:

$4 \left[\setminus \frac{1}{4} {x}^{2} = 3 x - \setminus \frac{25}{2}\right]$

${x}^{2} = 12 x - 50$

${x}^{2} - 12 x + 50 = 0$

If we check the discriminant of the form $a {x}^{2} + b x + c$:

${b}^{2} - 4 a c$

${\left(- 12\right)}^{2} - 4 \cdot 1 \cdot 50 = - 56$

Since the discriminant is negative there are no real solutions to this polynomial as can be seen from its graph:

graph{x^2-12x+50 [-35.17, 44.83, -3.52, 36.48]}