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QuestionSolve for real solutions using the quadratic formula: 4y2−y+4=04 y^{2}-y+4=0.

Studdy Solution

STEP 1

Assumptions1. The equation is quadratic in nature, i.e., it is of the form ax^ + bx + c =0. The coefficients a, b, and c are real numbers3. The quadratic formula is given byx=−b±b−4acax = \frac{-b \pm \sqrt{b^ -4ac}}{a}4. The equation given is4y−y+4=04y^ - y +4 =0

STEP 2

Identify the coefficients a, b, and c from the given equation.In this case, a =4, b = -1, and c =4.

STEP 3

Substitute the values of a, b, and c into the quadratic formula.
y=−(−1)±(−1)2−∗∗2∗y = \frac{-(-1) \pm \sqrt{(-1)^2 -**}}{2*}

STEP 4

implify the equation.
y=1±1−648y = \frac{1 \pm \sqrt{1 -64}}{8}

STEP 5

Further simplify the equation.
y=1±−638y = \frac{1 \pm \sqrt{-63}}{8}

STEP 6

The square root of a negative number is not a real number, it is an imaginary number. Therefore, the equation has no real solutions.

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