Question

write zeroes of x^2-x-6 and 4x^2+5√2x-3​

Answer 1

Step-by-step explanation:

$\huge \: 1) {x}^{2} - x - 6 \\ \\ {x}^{2} - x - 6 = 0 \\ \\ {x}^{2} + 2x - 3x - 6 = 0 \\ \\ x(x + 2) - 3(x + 2) = 0 \\ \\ (x - 3)(x + 2) = 0 \\ \\ x - 3 = 0 \: \: \: \: or \: \: \: x + 2 = 0 \\ \\ x = 3 \: \: \: \: or \: \: \: \: x = - 2 \\ \\ \huge \: 2)4 {x}^{2} + 5 \sqrt{2} x - 3 \\ \\ 4 {x}^{2} + 5 \sqrt{2} x - 3 = 0 \\ \\ a = 4 \\ b = 5 \sqrt{2} \\ c = - 3 \\ \\ \delta = \frac{ {b}^{2} - 4ac}{2a} \\ \\ \delta = \frac{( {5 \sqrt{2}) }^{2} - 4(4)( - 3) } {2(4)} \\ \\ \delta = \frac{50 + 48}{8} \\ \\ \delta = \frac{98}{8} \\ \\ \delta = \frac{49 \times 2}{4 \times 2} = \frac{49}{4} \\ \\ x = \frac{ - b \pm \sqrt{ \delta} }{2a} \\ \\ x = \frac{ - 5 \sqrt{2} \pm \sqrt{ \frac{49}{4} } }{2(4)} \\ \\ x = \frac{ - 5 \sqrt{2} \pm \frac{7}{2} }{8} \\ \\ x = \frac{ - 10 \sqrt{2} \pm7 }{8 \times 2} = \frac{ - 10 \sqrt{2} \pm7 }{16} \\ \\ x = - \frac{10}{16} \sqrt{2} + \frac{7}{16} \: \: \: \: or \: \: \: x = - \frac{10}{16} \sqrt{2} - \frac{7}{16}$

Answer 2

$\bold{{x}^{2} - x - 6}$

To find the zeroes

p(x) = 0

$p(x) = {x}^{2} - x - 6 = 0 \\ \implies {x}^{2} +2x -3x - 6 = 0 \\ \implies x(x+2) -3(x+2) = 0 \\ \implies (x-3) (x+2) = 0 \\ \implies (x-3) = 0 \; ; (x+2) = 0$

x - 3 = 0

x = 3

x + 2 = 0

x = -2

Zeroes of $p(x) = {x}^{2} - x - 6$ are 3 and -2

$\bold{4{x}^{2} + 5\sqrt{2}x - 3}$

To find the zeroes

f(x) = 0

$f(x) = 4{x}^{2} + 5\sqrt{2}x - 3 = 0 \\ \implies 4{x}^{2} + 6\sqrt{2}x -\sqrt{2}x - 3 \\ \implies 2\sqrt{2}x(\sqrt{2}x + 3) -1(\sqrt{2}x +3) \\ \implies (2\sqrt{2}x -1)(\sqrt{2} +3) = 0 \\ \implies (2\sqrt{2}x -1) = 0 \; ; (\sqrt{2} +3) = 0$

$2\sqrt{2}x -1 = 0 \\ x = \dfrac{1}{2\sqrt{2}}$

$\sqrt{2}x +3 = 0 \\ x = \dfrac{-3}{\sqrt{2}}$

Zeroes of $f(x) = 4{x}^{2} + 5\sqrt{2}x - 3$ are $\dfrac{1}{2\sqrt{2}}$ and $\dfrac{-3}{\sqrt{2}}$