SAT Nonlinear equations · one variable

Using the quadratic formula or factoring $2([MATH]x^{2} + 4x - 5)$ = 2 (x + 5) (x - 1) = 0[/MATH] gives solutions $x = -5$ and $x = 1$. Their sum is -5 + 1 = -4.

Expanding gives $x^{2}$ − $8x + 16 = 25$, or $x^{2}$ − 8x − 9 = 0. The discriminant is (−8)$^{2}$ − 4(1)(−9) = 64 + 36 = 100. Since the discriminant is positive, the equation has exactly two distinct real solutions.

For a quadratic equation to have exactly one solution, the discriminant must equal zero. For $x^{2}$ + bx + 9 = 0, the discriminant is $b^{2}$ - 4(1)(9) = $b^{2}$ - 36. Setting $b^{2}$ - 36 = 0 gives $b^{2}$ = 36, so $b = 6$ or $b = -6$. Since the choices include only positive values, $b = 6$.

Factoring the equation yields $(2x - 3) (3x + 5) = 0$, giving solutions $x = 3/2$ and $x = -5/3$. The positive solution is 3/2.

Squaring both sides yields $2x + 5 = (x - 5)^{2}$, which expands to $2x + 5$ = $x^{2}$ - $10x + 25$. Rearranging gives $x^{2}$ - $12x + 20 = 0$, which factors as $(x - 2) (x - 10) = 0$. The solutions are $x = 2$ and $x = 10$. Checking $x = 2$: $\displaystyle \sqrt{9} = 3$ but 2 - 5 = -3, so $x = 2$ is extraneous. Checking $x = 10$: $\displaystyle \sqrt{25} = 5$ and 10 - 5 = 5, so $x = 10$ is valid.