Question

Plz solve it
Std 10th Maharashtra board​

Answer 1

bring the equation to the form ax²+bx+c=0, where a, b, and c are coefficients. plug these coefficients in the formula: (-b±√(b²-4ac))/(2a)

Q. 1

a)

${x}^{2} - 4x - 3 = 0$

Here

a=1,b=-4,c=-3

Putting values in the formula

$\frac{-b±√(b²-4ac)}{2a} \\ = \frac{ - ( - 4)± \sqrt{ {( - 4)}^{2} - 4 \times 1 \times ( - 3)} }{2 \times 1} \\ = \frac{4 ± \sqrt{16 + 12} }{2} \\ = \frac{4 ± \sqrt{28} }{2} \\ = \frac{4 ± \sqrt{4 \times 7} }{2} \\ = \frac{4 ± 2 \sqrt{7} }{2} \\ we \: have \: the \: roots \\ 2 + \sqrt{7} \: and \: 2 - \sqrt{7}$

b)

$2 {x}^{2} + 5x - 2 = 0$

Here,

a=2,b=5,c=-2

Putting values in the formula

$\frac{-b±√(b²-4ac)}{2a} \\ = \frac{ - 5± \sqrt{ {5}^{2} - 4 \times 2 \times ( - 2) } }{2 \times 2} \\ = \frac{ - 5± \sqrt{ 25 + 16 } }{4} \\ = \frac{ - 5± \sqrt{ 41} }{4}$

Roots are

$\frac{ - 5 + \sqrt{ 41} }{4} \: and \: \frac{ - 5 - \sqrt{ 41} }{4}$

Q. 2

a)

${x}^{2} + 2x - 24\\ {x}^{2} - 4x + 6x - 24 \\ x(x - 4) + 6(x - 4) \\ (x + 6)(x - 4)$

b)

$3 {x}^{2} + 5x + 2 \\ 3 {x}^{2} + x + 2x + 2 \\ 3x(x + 1) + 2(x + 1) \\ (3x + 2)(x + 1)$

c)

$2 {x}^{2} - 5x - 12 \\ 2 {x}^{2} - 8x + 3x - 12 \\ 2x(x - 4) + 3(x - 4) \\ (2x + 3)(x - 4)$