Asked by Horacio Nicolas on Apr 27, 2024
Verified
Use the Quadratic Formula to solve t2+5t+6=0t ^ { 2 } + 5 t + 6 = 0t2+5t+6=0 .
A) t=2,t=3t = 2 , t = 3t=2,t=3
B) t=−2,t=3t = - 2 , t = 3t=−2,t=3
C) t=2,t=−3t = 2 , t = - 3t=2,t=−3
D) t=−2,t=−3t = - 2 , t = - 3t=−2,t=−3
E) no solutions
Quadratic Formula
A formula that provides the solutions to a quadratic equation in terms of its coefficients.
- Use the quadratic formula as a method to solve for the variables in quadratic equations.
Verified Answer
YA
yusof asefiApr 30, 2024
Final Answer :
D
Explanation :
The quadratic formula states that for a quadratic equation in the form of $ax^2+bx+c=0$, the solutions for $x$ are given by: x=−b±b2−4ac2ax = \frac{-b \pm \sqrt{b^2-4ac}}{2a}x=2a−b±b2−4ac
In this case, $a=1$, $b=5$, and $c=6$. Substituting those values into the formula gives:
t=−5±52−4(1)(6)2(1)t = \frac{-5 \pm \sqrt{5^2-4(1)(6)}}{2(1)}t=2(1)−5±52−4(1)(6)
Simplifying gives:
t=−5±12t = \frac{-5 \pm \sqrt{1}}{2}t=2−5±1
Therefore, the solutions are $t=-2$ and $t=-3$, which corresponds to choice D.
In this case, $a=1$, $b=5$, and $c=6$. Substituting those values into the formula gives:
t=−5±52−4(1)(6)2(1)t = \frac{-5 \pm \sqrt{5^2-4(1)(6)}}{2(1)}t=2(1)−5±52−4(1)(6)
Simplifying gives:
t=−5±12t = \frac{-5 \pm \sqrt{1}}{2}t=2−5±1
Therefore, the solutions are $t=-2$ and $t=-3$, which corresponds to choice D.
Learning Objectives
- Use the quadratic formula as a method to solve for the variables in quadratic equations.