What are the solutions to the equation 64x^2-(16a+4b)x+ab=0 ?

The solutions to the equation 64x^2-(16a+4b)x+ab=0 are x= a/4 ,x= b/(16)

Find the zeros of 64x^2-(16a+4b)x+ab=0

The zeros of 64x^2-(16a+4b)x+ab=0 are x= a/4 ,x= b/(16)

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64 x^{2} - (16 a + 4 b) x + ab = 0

x = \frac{a}{4}, x = \frac{b}{16}

Solution steps

64 x^{2} - (16 a + 4 b) x + ab = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 16 a - 4 b ) \pm ( - 16 a - 4 b ) ^{2} - 4 \cdot 64 ab}{2 \cdot 64}

Simplify

(- 16 a - 4 b)^{2} - 4 \cdot 64 ab : 4 (4 a - b)

x_{1, 2} = \frac{- ( - 16 a - 4 b ) \pm 4 ( 4 a - b )}{2 \cdot 64}

Separate the solutions

x_{1} = \frac{- ( - 16 a - 4 b ) + 4 ( 4 a - b )}{2 \cdot 64}, x_{2} = \frac{- ( - 16 a - 4 b ) - 4 ( 4 a - b )}{2 \cdot 64}

x = \frac{- ( - 16 a - 4 b ) + 4 ( 4 a - b )}{2 \cdot 64} : \frac{a}{4}

x = \frac{- ( - 16 a - 4 b ) - 4 ( 4 a - b )}{2 \cdot 64} : \frac{b}{16}

The solutions to the quadratic equation are:

x = \frac{a}{4}, x = \frac{b}{16}

14 + 3 h = 5 hh

y, x^{2} + x y + y^{2} = 3

x, x^{2} + y^{2} = a^{2}

y, x^{2} + 5 x y - 4 y^{2} = 10

x^{2} - x - 1

What are the solutions to the equation 64x^2-(16a+4b)x+ab=0 ?
The solutions to the equation 64x^2-(16a+4b)x+ab=0 are x= a/4 ,x= b/(16)
Find the zeros of 64x^2-(16a+4b)x+ab=0
The zeros of 64x^2-(16a+4b)x+ab=0 are x= a/4 ,x= b/(16)