Question

Using the Completing the Square method, what are the zeros of the quadratic function f(x) = 2x² + 8x-3?

Answer 1

Answer:

$x = -2 - \sqrt{\frac{11}{2}$ and $x = -2 + \sqrt{\frac{11}{2}}$

Step-by-step explanation:

Hello!

First factor out the coefficient of x² from the entire expression.

• $f(x) = 2x^2 + 8x - 3$
• $f(x) = 2(x^2 + 4x -\frac32)$

Standard form of a quadratic: $ax^2 + bx + c = 0$

Vertex form (completing the square): $y = a(x - h)^2 + k$

Perfect Square trinomial: $(a+b)^2 = a^2 + 2ab + b^2$

Given the equation in the brackets: $x^2 + 4x - \frac32$

• a = 1
• b = 4
• c = -3/2

We want to convert the parentheses into a Perfect Square Trinomial (PST).

To find the missing term to complete the square, we must:

• take the b-value (4)
• divide it by two (4/2 = 2)
• square it (2^2 = 4)

To change -3/2 to 4, we have to add 5 and 1/2.

But we also want to balance our equation, as we don't want to change the value of the equation. Since we are adding 5.5, and it is being multiplied by 2, we want to subtract 2(5.5) or 11.

• $f(x) = 2(x^2 + 4x -\frac32)$
• $f(x) = 2(x^2 + 4x -\frac32 + 5 \frac12) - 2(5 \frac12)$
• $f(x) = 2(x^2 + 4x + 4) - 11$

Convert the parentheses into factored form.

• $f(x) = 2(x + 2)^2 - 11$

Set the equation to 0 and use the square root property:

• $f(x) = 2(x + 2)^2 - 11$
• $0 = 2(x + 2)^2 - 11$
• $11 = 2(x + 2)^2$
• $\frac{11}{2} = (x + 2)^2$
• $\sqrt{\frac{11}{2} = (x + 2)^2}$
• $\pm\sqrt{\frac{11}{2}} = x + 2$
• $-2\pm\sqrt{\frac{11}{2}} = x$

The zeroes are $x = -2 - \sqrt{\frac{11}{2}$ and $x = -2 + \sqrt{\frac{11}{2}}$.