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3 years ago

Help me solve this please. The bottom one

Mathematics

1 answer:

Nesterboy [21]3 years ago

7 0

The answers is c......

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Quadratic Equation

Natasha_Volkova [10]

Step-by-step explanation:

The sum of ages of two friends is 13 years.

The product of their ages is 42.

Let the age of 1st friend and 2nd friend is x, y respectively.

1 st condition= The sum of ages of two friends is 13 yrs.

i.e x+y = 13........ (I)

2nd condition= The product of their ages is 42.

i.e X*y = 42........(ii)

From equation (I)

X+y = 13

or, X = 13-y........ (iii)

Putting the equation (iii) in equation (ii).

X*y= 42

(13-y) * y = 42

13y - y^2 = 42

${y}^{2} - 13y + 42 = 0$

${y}^{2} - (7 + 6)y + 42= 0$

${y}^{2} - 7y - 6y + 42 = 0$

$y(y - 7) - 6(y - 7) = 0$

$(y - 6) (y - 7) = 0$

Either; y-6 = 0

y = 6

Or;

y-7=0

y = 7

Keeping the value of y as "7" in equation (ii)

x*y = 42

7x = 42

X = 42/7

Therefore, the value of X is 6.

Therefore, either 1st friend is 6 years and 2nd is 7 years.

Hope it helps...

7 0

3 years ago

The area of the triangle is eqaula to the sqare show that x²-3x-2=0

agasfer [191]

Answer:

Step-by-step explanation:

Area of square = area of triangle

$side*side = \dfrac{1}{2}b*h\\\\\\x*x=\dfrac{1}{2}*(x+1)(x+2)\\x^{2}=\dfrac{1}{2}[x*x +x*2+1*x+1*2]\\\\\\x^{2}=\dfrac{1}{2}[x^{2}+2x+x+2]\\\\\\x^{2}=\dfrac{1}{2}[x^{2}+3x+2]\\\\\\Multiply \ both \ sides \ by \ 2\\\\2x^{2}=x^{2}+3x+2\\\\2x^{2}-x^{2}-3x-2 = 0$