Question

Verify that the roots of 5x²- 6x -2 = 0 are 0" id="TexFormula1" title="\frac{3 + \sqrt{19} }{5} " alt="\frac{3 + \sqrt{19} }{5} " align="absmiddle" class="latex-formula"> and $\frac{3 - \sqrt{19} }{5}$​

Answer 1

Since, LHS = RHS = 0, They are the roots of equation.

Answer 2

Answer:

Proof below.

Step-by-step explanation:

Quadratic Formula

$x=\dfrac{-b \pm \sqrt{b^2-4ac} }{2a}\quad\textsf{when }\:ax^2+bx+c=0$

Given quadratic equation:

$5x^2-6x-2=0$

Define the variables:

• a = 5
• b = -6
• c = -2

Substitute the defined variables into the quadratic formula and solve for x:

$\implies x=\dfrac{-(-6) \pm \sqrt{(-6)^2-4(5)(-2)}}{2(5)}$

$\implies x=\dfrac{6 \pm \sqrt{36+40}}{10}$

$\implies x=\dfrac{6 \pm \sqrt{76}}{10}$

$\implies x=\dfrac{6 \pm \sqrt{4 \cdot 19}}{10}$

$\implies x=\dfrac{6 \pm \sqrt{4}\sqrt{19}}{10}$

$\implies x=\dfrac{6 \pm2\sqrt{19}}{10}$

$\implies x=\dfrac{3 \pm \sqrt{19}}{5}$

Therefore, the exact solutions to the given quadratic equation are:

$x=\dfrac{3 + \sqrt{19}}{5} \:\textsf{ and }\:x=\dfrac{3 - \sqrt{19}}{5}$

Learn more about the quadratic formula here:

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