Math

QuestionSolve the equation 4w2+3w+12=24 w^{2}+3 w+12=2 and simplify your answer, including non-real solutions. w= w=

Studdy Solution

STEP 1

Assumptions1. We are solving for the variable ww in the quadratic equation 4w+3w+12=4w^ +3w +12 =. . The solutions can be real or non-real.

STEP 2

First, we need to rearrange the equation to standard quadratic form ax2+bx+c=0ax^2 + bx + c =0. We can do this by subtracting2 from both sides of the equation.
4w2+w+122=224w^2 +w +12 -2 =2 -2

STEP 3

implify the equation.
w2+3w+10=0w^2 +3w +10 =0

STEP 4

The quadratic formula is given by w=b±b24ac2aw = \frac{-b \pm \sqrt{b^2 -4ac}}{2a}. We can use this formula to solve for ww.

STEP 5

Identify the values of aa, bb, and cc from the equation 4w2+3w+10=04w^2 +3w +10 =0.
a=4a =4, b=3b =3, and c=10c =10

STEP 6

Substitute the values of aa, bb, and cc into the quadratic formula.
w=3±32441024w = \frac{-3 \pm \sqrt{3^2 -4 \cdot4 \cdot10}}{2 \cdot4}

STEP 7

implify the expression under the square root.
w=3±9160w = \frac{-3 \pm \sqrt{9 -160}}{}

STEP 8

Further simplify the expression.
w=3±1518w = \frac{-3 \pm \sqrt{-151}}{8}

STEP 9

Since the number under the square root is negative, the solutions will be non-real. We can express the square root of a negative number in terms of ii, where i2=i^2 = -.
w=3±i1518w = \frac{-3 \pm i\sqrt{151}}{8}The solutions to the quadratic equation 4w2+3w+=4w^2 +3w + = are w=3+i1518w = \frac{-3 + i\sqrt{151}}{8} and w=3i1518w = \frac{-3 - i\sqrt{151}}{8}.

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