How do you solve the quadratic using the quadratic formula given $4 x^{2} + 4 x - 8 = 1$ $4x^2+4x-8=1$ ?

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How do you solve the quadratic using the quadratic formula given $4 x^{2} + 4 x - 8 = 1$ $4x^2+4x-8=1$ ?

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Answer 1 · 2016-08-29T14:57:44

Solution is $\frac{- 1 - \sqrt{10}}{2}$ $(-1-sqrt10)/2$ and $\frac{- 1 + \sqrt{10}}{2}$ $(-1+sqrt10)/2$ .

Explanation:

For the equation $a x^{2} + b x + c = 0$ $ax^2+bx+c=0$ , the roots are given by $x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$ $x=(-b+-sqrt(b^2-4ac))/(2a)$ . As can be seen type of roots depend upon discriminant $b^{2} - 4 a c$ $b^2-4ac$ . If is negative, roots are complex and if it is a complete square (and coefficients are rational), roots are rational.

In the equation $4 x^{2} + 4 x - 8 = 1$ $4x^2+4x-8=1$ or $4 x^{2} + 4 x - 9 = 0$ $4x^2+4x-9=0$ , the discriminant is ${(4)}^{2} - 4 \times 4 \times (- 9) = 16 + 144 = 160$ $(4)^2-4×4×(-9)=16+144=160$ , which is not a complete square and hence roots are irrational.

Roots are $x = \frac{- 4 \pm \sqrt{160}}{2 \times 4}$ $x=(-4+-sqrt160)/(2×4)$

= $\frac{- 4 \pm 4 \sqrt{10}}{8}$ $(-4+-4sqrt10)/8$

= $\frac{- 1 \pm \sqrt{10}}{2}$ $(-1+-sqrt10)/2$

Hence solution is $\frac{- 1 - \sqrt{10}}{2}$ $(-1-sqrt10)/2$ and $\frac{- 1 + \sqrt{10}}{2}$ $(-1+sqrt10)/2$ .

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How do you solve the quadratic using the quadratic formula given $4 x^{2} + 4 x - 8 = 1$ $4x^2+4x-8=1$ ?

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