Question

$5x {}^{2} - 4x - 3 = 0 \: solve \: using \: formula \: method$
​

Answer 1

Given Equation

$\tt : \implies \: 5 {x}^{2} - 4x - 3 = 0$

To Find

$\tt : \implies \: zeroes \:$

Formula

$\tt : \implies \: x = \dfrac{ - b \pm \sqrt{ {b}^{2} - 4ac } }{2a}$

Now Compare with

$\tt : \implies \: {ax}^{2} + bx + c = 0$

we get

$\tt : \implies \: a = 5,b = - 4 \: and \: c \: = - 3$

Now put the value on formula

$\tt : \implies\: x = \dfrac{ - ( - 4) \pm \sqrt{( - 4) ^{2} - 4 \times 5 \times - 3} }{2 \times 5}$

$\tt : \implies x = \dfrac{4 \pm \sqrt{16 + 20 \times 3} }{10}$

$\tt : \implies \: x = \dfrac{4 \pm \sqrt{16 + 60} }{10}$

$\tt : \implies \: x = \dfrac{4 \pm \sqrt{76} }{10}$

$\tt : \implies \: x = \dfrac{4 \pm 2\sqrt{19} }{10}$

$\tt : \implies \: x = \dfrac{2 \pm \sqrt{19} }{5}$

$\tt : \implies \: x \: = \dfrac{2 + \sqrt{19} }{5} \: \: and \: x = \dfrac{2 - \sqrt{19} }{5}$

Answer

$\tt : \implies \: x \: = \dfrac{2 + \sqrt{19} }{5} \: \: and \: x = \dfrac{2 - \sqrt{19} }{5}$

Answer 2

${\pmb{\sf{\bigstar \: {\underline{Required \; Solution...}}}}}$

Solve the following by using formula: 5x²-4x-3=0, have to find zeroes of the given equation. Let's do it!

${\pmb{\sf{\bigstar \: {\underline{Using \; formula...}}}}}$

${\small{\underline{\boxed{\sf{\bull \: x \: = \dfrac{-b+\sqrt{b^2-4ac}}{2a}}}}}}$

${\pmb{\sf{\bigstar \: {\underline{Firstly, \: Some \: Knowledge...}}}}}$

Some other knowledge about Quadratic Equations -

★ Sum of zeros of any quadratic equation is given by ➝ α+β = -b/a

★ Product of zeros of any quadratic equation is given by ➝ αβ = c/a

★ Discriminant is given by b²-4ac

• Discriminant tell us about there are solution of a quadratic equation as no solution, one solution and two solutions.

★ A quadratic equation have 2 roots

★ ax² + bx + c = 0 is the general form of quadratic equation.

⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━

${\pmb{\sf{\bigstar \: {\underline{Full \; Solution...}}}}}$

~ Firstly let us compare by using the general form of quadratic equation that is ax² + bx + c = 0 we get the following:

• a is 5, b is -4 and c is -3

~ Now by using the formula let us solve the question!

${\small{\underline{\boxed{\sf{\bull \: x \: = \dfrac{-b+\sqrt{b^2-4ac}}{2a}}}}}} \\ \\ :\implies \sf x \: = \dfrac{-b+\sqrt{b^2-4ac}}{2a} \\ \\ :\implies \sf x \: = \dfrac{-(-4) + \sqrt{(-4)^2 - 4(5)(-3)}}{2(5)} \\ \\ :\implies \sf x \: = \dfrac{-(-4) + \sqrt{(-4)^2 - 4 \times 5 \times -3}}{2(5)} \\ \\ :\implies \sf x \: = \dfrac{-(-4) + \sqrt{(-4)^2 - 4 \times -15}}{2(5)} \\ \\ :\implies \sf x \: = \dfrac{-(-4) + \sqrt{-4 \times -4 - 4 \times -15}}{2(5)} \\ \\ :\implies \sf x \: = \dfrac{-(-4) + \sqrt{ - 4 \times -4 - 4 \times -15}}{2(5)} \\ \\ :\implies \sf x \: = \dfrac{-(-4) + \sqrt{16 + 60}}{2 \times 5} \\ \\ :\implies \sf x \: = \dfrac{-(-4) + \sqrt{76}}{10} \\ \\ :\implies \sf x \: = \dfrac{4+2 \sqrt{19}}{10} \\ \\ :\implies \sf x \: = \dfrac{2+\sqrt{19}}{5} \: or \: x \: = \dfrac{2-\sqrt{19}}{5}$