Question

solve using the quadratic formula: 5x² + 3x - 1 = 0
x = (3 ± √29)/5
x = (-3 ± √29)/10
solve by factoring. x² + 5x - 36 = 0
solve the following equation using the quadratic formula: 3x² + 16x + 5 = 0
when 3x + 2 ≤ 5(x - 4) is solved for x, the solution is 1) x ≤ 3 2) x ≥ 3 3) x ≤ -11 4) x ≥ 11
what is the solution set for the equation |3x - 1| = x + 5? 1) {-1} 2) {-1,3} 3) {3} 4) {1,-3}
what is the solution to -3(x - 6) > 2x - 2? 1) x > 4 2) x < 4 3) x > -16 4) x < -16

Explanation:

Step1: Identify coefficients for $5x^{2}+3x - 1=0$

For a quadratic equation $ax^{2}+bx + c = 0$, here $a = 5$, $b = 3$, $c=-1$.

Step2: Apply quadratic formula $x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$

$x=\frac{-3\pm\sqrt{3^{2}-4\times5\times(-1)}}{2\times5}=\frac{-3\pm\sqrt{9 + 20}}{10}=\frac{-3\pm\sqrt{29}}{10}$

Step3: Factor $x^{2}+5x - 36=0$

We need two numbers that multiply to - 36 and add up to 5. The numbers are 9 and - 4. So, $(x + 9)(x-4)=0$. Then $x=-9$ or $x = 4$.

Step4: Identify coefficients for $3x^{2}+16x + 5=0$

Here $a = 3$, $b = 16$, $c = 5$.

Step5: Apply quadratic formula

$x=\frac{-16\pm\sqrt{16^{2}-4\times3\times5}}{2\times3}=\frac{-16\pm\sqrt{256-60}}{6}=\frac{-16\pm\sqrt{196}}{6}=\frac{-16\pm14}{6}$. We get $x=\frac{-16 + 14}{6}=-\frac{1}{3}$ and $x=\frac{-16-14}{6}=-5$.

Step6: Solve $3x + 2\leq5(x - 4)$

Expand: $3x+2\leq5x-20$. Move terms: $2 + 20\leq5x-3x$, $22\leq2x$, so $x\geq11$.

Step7: Solve $|3x - 1|=x + 5$

Case 1: $3x-1=x + 5$, then $3x-x=5 + 1$, $2x=6$, $x = 3$. Case 2: $3x-1=-(x + 5)$, $3x-1=-x-5$, $3x+x=-5 + 1$, $4x=-4$, $x=-1$. Check: when $x = 3$, $|3\times3-1|=|8|=8$ and $3 + 5=8$; when $x=-1$, $|3\times(-1)-1|=| - 4|=4$ and $-1+5=4$. So the solution set is $\{-1,3\}$.

Step8: Solve $-3(x - 6)>2x-2$

Expand: $-3x+18>2x-2$. Move terms: $18 + 2>2x+3x$, $20>5x$, so $x<4$.

Answer:

1. $x=\frac{-3\pm\sqrt{29}}{10}$
2. $x=-9$ or $x = 4$
3. $x=-\frac{1}{3}$ or $x=-5$
4. $x\geq11$
5. $\{-1,3\}$
6. $x<4$