Question

2x^2 + √7x -7 = 0 solve quadratic equation by formula

Answer 1

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Answer 2

Answer:

The answer is $\sqrt{7} /2$ and $-\sqrt{7} / 2$.

Step-by-step explanation:

A quadratic equation exists as an algebraic equation of the second degree in x. The quadratic equation in its standard form exists $ax^{2} + bx + c = 0,$ where a and b stand as the coefficients, x exists as the variable, and c is the constant term. The first condition for an equation to be a quadratic equation stands for the coefficient of $x^{2}$ exists as a non-zero term(a ≠0).

Given,

$2 x^{2}+\sqrt{7} x-7=0$

To solve the equation,

$2 x^{2}+\sqrt{7} x-7=0$

Where,

a=2

b=$\sqrt{7}$

c=-7

$D=b^{2}-4 a c$

Putting the values in the equation,

$=(\sqrt{7})^{2}-4 \times 2 \times(-7)$

$=7+56$

$=63$

$\sqrt{D}=\sqrt{63}$

$\alpha=\frac{-b+\sqrt{D}}{2 a}=\frac{\sqrt{7}+\sqrt{63}}{4}$

$=\frac{\sqrt{7}(3-1)}{4}$

Simplifying,

$=\frac{2 \sqrt{7}}{42}$

We get,

$=\sqrt{7} /2$

Or

$\beta=-\frac{b-\sqrt{D}}{2 a} &=\frac{-\sqrt{7}-\sqrt{63}}{4}$

$&=\frac{-\sqrt{7}(3-1)}{4}$

Then,

$&=\frac{-2 \sqrt{7}}{4 / 2}$

We get,

$&=-\sqrt{7} / 2$

Therefore, The answer is $\sqrt{7} /2$ and $-\sqrt{7} / 2$.

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