Question

factorise given quadratic equation .

x² +2x-5​

Answer 1

Answer:

x² + 2x - 5 = 0

So, we have to find to it's zeroes by using completing Square method.

x² + 2x - 5 = 0

we can write 2x = 2(x)(1)

x² + (2)(x)(1) + [ 1² - 1²] - 5 = 0

(x + 1)² - 1 - 5 = 0

[(x² + 2x + 1) = (x + 1)² , where, (a² + b² + 2ac = (a + b)²]

so,

(x + 1)² = 6

[taking Square Root both side]

(x + 1) = ±√6

now there is two situation

( 1 ) we have to take (+ve) sign.

So, answeer will be.

(x + 1) = √6

x = (√6 - 1)

( 2 ) we have to take (+ve) sign.

So, answeer will be.

(x + 1) = -√6

x = -(√6 + 1)

Step-by-step explanation:

hope its helpful

Answer 2

$x {}^{2} + 2x - 5 = 0 \\ \\ = > x {}^{2} + 2x = 5 \\ = > 4 (x {}^{2} + 2) = 5 \times 4 \\ = > 4x {}^{2} + 8 = 20 \\ = > 4x + 8 + 2 {}^{2} = 20 + 2 { }^{2} \\ = > 4x { }^{2} + 8 + 4 = 20 + 4 \\ = > 4x {}^{2} + 12 = 24 \\ = > 4x {}^{2} = 24 - 12 \\ = > 4x {}^{2} = 12 \\ = > x {}^{2} = \frac{12}{4 } \\ = > x {}^{2} = 3 \\ = > x = \sqrt{3}$