if p and q are roots of the equation x^2-7x+10=0 then find the equation whose roots are (p+2) (q+2)
Answers
Answer :- x² - 11x + 28 = 0
Given :-
◉ Quadratic equation : x² - 7x + 10 = 0, where p and q are the zeroes of it.
To Find :-
◉ A quadratic equation whose zeroes are (p + 2) and (q + 2)
Solution :-
Let's find the value of p and q, at first.
⇒ x² - 7x + 10 = 0
⇒ x² - 5x - 2x + 10 = 0
⇒ x(x - 5) - 2(x - 5) = 0
⇒ (x - 5)(x - 2) = 0
Now, We have found the value of p and q, So
we now need to find the Quadratic Equation whose zeroes are (p + 2)(q + 2)
⇒ p + 2 ⇒ 5 + 2 ⇒ 7
⇒ q + 2 ⇒ 2 + 2 ⇒ 4
Now,
A quadratic equation = x² - (sum of zeroes)x + (product of zeroes)
⇒ x² - (7 + 4)x + (7×4)
⇒ x² - 11x + 28
Hence, The required quadratic equation is x² - 11x + 28 = 0
More Information :-
◉ A quadratic equation is expressed as x² - (sum of zeroes)x + (product of zeroes)
Also,
◉ Sum of zeroes = ( - coefficient of x ) / (coefficient of x² )
◉ Product of zeroes = ( constant term ) / (coefficient of x² )
◉ The discriminant of a quadratic equation can be used to know the nature of roots. D = b² - 4ac
When,
- D = 0
Two real and equal roots exist.
- D < 0
No real roots exit, but two imaginary roots.
- D > 0
Two real and distinct roots exist.
★Answer:–
equation - x²- 7x + 10
if p and q are the roots of this equation
then , p + q = -(-7) = 7 ( equation 1 )
pq = 10 ( equation 2 )
the zeroes of the new equation = p + 2 , q + 2
sum of zeroes of new equation = p + q + 4
= 7 + 4 = 11 ( p + q = 7 from equation 1 )
product of zeroes of new equation = pq + 2 ( p + q ) + 4
= 10 + 2(7) + 4
= 10 + 14 + 4 = 28
any quadratic equation can be expressed as
=x²- ( sum of zeroes ) x + ( product of zeroes )
=x²-(11)x+28
=x²-11x+28
the new quadratic equation is x² - 11x + 28