Solve the equation by the quadratic formula 3×2 + 8x – 16 = 0 X= (Type an exact answer; using radicals as needed. Express
The Correct Answer and Explanation is:
We are given the quadratic equation:3×2+8x−16=03x^2 + 8x – 16 = 03×2+8x−16=0
Step 1: Identify coefficients
Compare the equation with the standard form ax2+bx+c=0ax^2 + bx + c = 0ax2+bx+c=0.
Here,
- a=3a = 3a=3
- b=8b = 8b=8
- c=−16c = -16c=−16
Step 2: Apply the quadratic formula
The quadratic formula is:x=−b±b2−4ac2ax = \frac{-b \pm \sqrt{b^2 – 4ac}}{2a}x=2a−b±b2−4ac
Substitute the values of aaa, bbb, and ccc:x=−8±82−4(3)(−16)2(3)x = \frac{-8 \pm \sqrt{8^2 – 4(3)(-16)}}{2(3)}x=2(3)−8±82−4(3)(−16)
Step 3: Simplify inside the square root
x=−8±64+1926x = \frac{-8 \pm \sqrt{64 + 192}}{6}x=6−8±64+192x=−8±2566x = \frac{-8 \pm \sqrt{256}}{6}x=6−8±256
Since the square root of 256 is 16:x=−8±166x = \frac{-8 \pm 16}{6}x=6−8±16
Step 4: Solve both solutions
First solution (using plus sign):x=−8+166=86=43x = \frac{-8 + 16}{6} = \frac{8}{6} = \frac{4}{3}x=6−8+16=68=34
Second solution (using minus sign):x=−8−166=−246=−4x = \frac{-8 – 16}{6} = \frac{-24}{6} = -4x=6−8−16=6−24=−4
Final Answer:
x=43,x=−4x = \frac{4}{3},\quad x = -4x=34,x=−4
Explanation
To solve any quadratic equation using the quadratic formula, you first need to identify the coefficients of the equation. In our equation 3×2+8x−16=03x^2 + 8x – 16 = 03×2+8x−16=0, we recognized aaa, bbb, and ccc by matching it to the general form. Using the quadratic formula guarantees a way to solve any quadratic equation, even when it cannot be factored easily.
Next, we substituted the values into the formula. One common mistake students make is not handling negative signs properly during substitution, especially in the discriminant, which is the expression under the square root: b2−4acb^2 – 4acb2−4ac. In our case, we carefully calculated 82−4(3)(−16)8^2 – 4(3)(-16)82−4(3)(−16) which simplifies to 256, a perfect square.
After computing the square root and simplifying the numerator, we split the expression into two cases using the plus and minus signs in the formula. This gives us the two possible values of xxx that satisfy the equation. The results, 43\frac{4}{3}34 and −4-4−4, are exact because we used radicals and simplified only when the square root had a whole number result.
The quadratic formula is reliable and useful for all quadratic equations, especially when factoring is difficult or when the solutions are irrational.
