Question

If sum of the squares of zeroes of the quadratic
polynomial p(x) = x² – 10x + 2k is 28, find the
value of k.​

Answer 1

Given : -

Quadratic equation

p ( x ) = x² - 10x + 2k

Sum of the squares of zeroes of the Quadratic polynomial is 28 .

Required to find : -

• Value of k ?

Conditions used : -

Here, conditions refers to the relationship between the zeroes and the coefficients of the quadratic polynomial .

So,

The relationship between the sum of the zeroes and the coefficients is ;

α + β = - coefficient of x/ coefficient of x²

Similarly,

The relationship between the product of the zeroes and the coefficients is ;

α.β = constant term/ coefficient of x²

Solution : -

Quadratic equation

p ( x ) = x² - 10x + 2k

Sum of the squares of zeroes of the Quadratic polynomial is 28 .

we need to find the value of k ?

So,

p ( x ) = x² - 10x + 2k

The standard form of a quadratic equation is ax² + bx + c = 0 .

Compare the standard form with the given quadratic equation .

Here,

• a = 1

• b = - 10

• c = 2k

Let's consider the zeroes as alpha ( α ) , beta ( β )

As we know that ;

The relation between the sum of the zeroes and the coefficients is;

α + β = - coefficient of x/ coefficient of x²

So,

⟹ α + β = - b/a

⟹ α + β = - ( - 10 )/1

⟹ α + β = 10/1

⟹ α + β = 10

Similarly,

The relationship between the product of the zeroes and the coefficients is ;

α.β = constant term/ coefficient of x²

So,

⟹ α.β = c/a

⟹ α.β = 2k/1

⟹ α.β = 2k

Since,

It is mentioned that ;

Sum of the squares of zeroes of the Quadratic polynomial is 28 .

α² + β² = 28

This implies ;

⟹ α + β = 10

Squaring on both sides

⟹ ( α + β )² = ( 10 )²

⟹ α² + β² + 2 ( α.β ) = 100

since,

α² + β² = 28

α.β = 2k

⟹ 28 + 2 ( 2k ) = 100

⟹ 2 ( 2k ) = 100 - 28

⟹ 2 ( 2k ) = 72

⟹ 4k = 72

⟹ k = 72/4

⟹ k = 18

Therefore,

Value of k = 18

Answer 2

Answer:

There's an easy formula for this in quadratic equations,

If there's a quadratic expression,

a.x square + b.x + c …….(1)

The roots of equation are given as,(there will be 2 roots implies if x is given a value of one of those 2 roots, the expression (1) will evaluate to 0

The roots are,

x = {-(b) + sqrt( b square - 4×a×c)}/(2×a)

And

x = {-(b) - sqrt( b square - 4×a×c)}/(2×a)

Following the above formula,

If -4 is a root of the equation then substituting x as -4 will result in expression becoming a zero,

=> x = -4 in your given equation

=> (-4) square - (-4) -(2×k +2) =0

=> 16 +4 -2×k -2=0

=> 18–2×k=0

=> 2×k = 18

=> k = 18/2

=>k =9