Question

2x^2+x-508=0 finds the roots of quadratic Eqaution using squaring method.​

Answer 1

Answer:

2x

2

+x−508

view step

ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right)

x_{1}

x_{2}

ax^{2}+bx+c=0

2x^{2}+x-508=0

2x

2

+x−508=0

view step

ax^{2}+bx+c=0

\frac{-b±\sqrt{b^{2}-4ac}}{2a}

±

x=\frac{-1±\sqrt{1^{2}-4\times 2\left(-508\right)}}{2\times 2}

x=

2×2

−1±

1

2

−4×2(−508)

view step

1

x=\frac{-1±\sqrt{1-4\times 2\left(-508\right)}}{2\times 2}

x=

2×2

−1±

1−4×2(−508)

view step

-42

x=\frac{-1±\sqrt{1-8\left(-508\right)}}{2\times 2}

x=

2×2

−1±

1−8(−508)

view step

-8-508

x=\frac{-1±\sqrt{1+4064}}{2\times 2}

x=

2×2

−1±

1+4064

view step

14064

x=\frac{-1±\sqrt{4065}}{2\times 2}

x=

2×2

−1±

4065

view step

22

x=\frac{-1±\sqrt{4065}}{4}

x=

4

−1±

4065

view step

x=\frac{-1±\sqrt{4065}}{4}

±-1\sqrt{4065}\approx 63.757352517

x=\frac{\sqrt{4065}-1}{4}

x=

4

4065

−1

view step

x=\frac{-1±\sqrt{4065}}{4}

±\sqrt{4065}\approx 63.757352517

-1

x=\frac{-\sqrt{4065}-1}{4}

x=

4

−

4065

−1

view step

ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right)

\frac{-1+\sqrt{4065}}{4}\approx 15.689338129

x_{1}

\frac{-1-\sqrt{4065}}{4}\approx -16.189338129

x_{2}

2x^{2}+x-508=2\left(x-\frac{\sqrt{4065}-1}{4}\right)\left(x-\frac{-\sqrt{4065}-1}{4}\right)

2x

2

+x−508=2(x−

4

4065

−1

)(x−

4

−

4065

−1

)

is true

Next step