Question

÷÷÷÷÷maths aryabhattas÷÷÷÷÷÷


answer the question attached


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Answer 1

$x^{2} +10x+25\\\\x^{2} +5x+5x+25\\\\x(x+5)+5(x+5)\\\\(x+5)(x+5)$

If the sum of quadratic equation is 0.

$x^{2} +10x+25=0\\\\x^{2} +5x+5x+25=0\\\\x(x+5)+5(x+5)=0\\\\(x+5)=0\\\\x=(-5)$

The value of x = (-5)

factorise 25 as 5*5,

Write 10x as 5x+5x.

Answer 2

$\huge\star{\tt{\underline{\underline{\red{Answer}}}}}\star$

x² + 10x + 25

This can be solved by Quadratic Formula

i.e

$\large{\frac{-b ± \sqrt{{b}^{2} - 4ac}}{2a}}$

Here in this Question

The Quadratic equation will have a,b,c value as (ax²+bx+c)

$\large{\boxed{a=1}}$

$\large{\boxed{b=10}}$

$\large{\boxed{c=25}}$

Now by applying these values in the formula

$\large{\frac{-10 ± \sqrt{{10}^{2} - 4 \times 1 \times 25}}{2 \times 1}}$

$\implies$$\large{\frac{-10 ± \sqrt{100-100}}{2}}$

$\implies$ $\large{\frac{-10}{2}}$

$\huge{\boxed{\boxed{x=-5}}}$