Question

find the roots of the equation x^2+2x-35=0 are​

Answer 1

${\huge{\pink{\underline{\underline{\purple{\mathbb{\mathcal{\bold{AnSwEr}}}}}}}}}$

${\large{\blue{\bold{\underline{Given:-}}}}}$

$- > x {}^{2} + 2x - 35 = 0$

${\large{\blue{\bold{\underline{To\:Find:-}}}}}$

▪︎ Roots if the given equation.

${\large{\blue{\bold{\underline{Now,}}}}}$

$- > {x}^{2} + 2x - 35 = 0 \\ \\ - > {x}^{2} + (7 - 5)x - 35 = 0 \\ (splitting \: middle \:term) \\ \\ - > {x}^{2} + 7x - 5x - 35 = 0 \\ \\ - > x(x + 7) - 5(x + 7) = 35 \\ \\ - > (x - 5) \times (x + 7) = 0 \\ \\ - > (x - 5) = 0 \: or \: (x + 7) = 0 \\ \\ - > x = 5 \: or \: x = - 7$

▪︎ Therefore the required roots of the given quadratic equation are 5 and -7.

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Answer 2

Step-by-step explanation:

Given:

${x}^{2} + 2x - 35 = 0$

To find: Solution of quadratic equation using quadratic formula.

Solution:

Quadratic formula:

$\boxed{\bold{\pink{x_{1,2} = \frac{ - b ± \sqrt{ {b}^{2} - 4ac } }{2a}}}}$

for standard quadratic equation given by

$a {x}^{2} + bx + c = 0,a\neq0$

Step 1: Write the value of a,b and c

Compare the given equation by standard equation

$a = 1 \\ b = 2 \\ c = - 35$

Step 2: Put these values in quadratic formula

$x_{1,2} = \frac{ - 2 ± \sqrt{4 - 4(1)( - 35)} }{2} \\ \\ x_{1,2} = \frac{ - 2 ± \sqrt{4 + 140} }{2} \\ \\ x_{1,2} = \frac{ - 2 ± \sqrt{144} }{2} \\ \\ x_{1,2} = \frac{ - 2 ± 12}{2}$

Step 3: Take (+) sign

$x_1 = \frac{ - 2 + 12}{2} \\ \\ x_1 = \frac{10}{2} \\ \\ x_1 = 5$

Step 4: Take (-) sign

$x_2 = \frac{ - 2 - 12}{2} \\ \\ x_2 = \frac{ - 14}{2} \\ \\ x_2 = - 7$

Final answer:

Value of x are

$\bold{\red{x = 5}} \\ \\ \bold{\green{x = - 7}}$

Hope it helps you.

To learn more:

1) Without finding the zeroes a and B of

polynomial p (x) = 6x^2- 13x + 6, find the

values of the following:(i) 1/alpha+1/b...

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

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