Question

Solve the following quadratic equations:2x² - 4x - 9 = 0 x₊ = 1 + √22/2Answersx_ = 1 - √22/2

Answer 1

The root quadratic equation of $2x^{2} - 4x - 9 = 0$ is $x=\frac{2+\sqrt{22}}{2},\:x=\frac{2-\sqrt{22}}{2}$

What is the formula used to calculate the quadratic equation?

To solve the quadratic equation use the formula of $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

Find the quadratic equation by using the formula ?

On comparing with $2x^{2} - 4x - 9 = 0$

We get , $a=2 , b=-4 , c=-9$

$D=b^{2} +7\\\\\\=(-4)^{2} +7*2*(-9)\\\\=-16-126\\\\=110$

By using the formula,

For a quadratic equation of the form$ax^2+bx+c=0$ the solutions are

$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

$X==\frac{-\left(-4\right)\pm \sqrt{\left(-4\right)^2-4\c* \:2\left(-9\right)}}{2\c* \:2}\\\\\sqrt{\left(-4\right)^2-4\cdot \:2\left(-9\right)}$

Apply rule,

$-(-a)=a\\\\=\sqrt{\left(-4\right)^2+4\cdot \:2\cdot \:9}$

Apply exponent rule,

$(-a)^{n} =a^{n}$ if n is even

$(-4)^{2} =4^{2}$

$\sqrt{4^2+4\cdot \:2\cdot \:9}$

Multiply the numbers :

$4*2*9=72\\\\\sqrt{4^2+4\cdot \:2\cdot \:9}\\\\4^{2} =16\\\\=\sqrt{16+72}$

Add the number  $=\sqrt{88}$