We will now try to answer the question raised regarding the relationship between zeroes
and coefficients of a quadratic polynomial.
`color(red)(=>"For this, let us take a quadratic polynomial,")` say `color(blue)(p(x) = 2x^2 – 8x + 6)`.
In Class IX, you have learnt how to factorise quadratic polynomials by splitting the middle term.
So, here we need to split the middle term ‘– 8x’ as a sum of two terms, whose product is` 6 × 2x^2 = 12x^2`.
So, we write
`color(blue)(2x^2 – 8x + 6) = 2x^2 – 6x – 2x + 6 = 2x(x – 3) – 2(x – 3)`
`" " = (2x – 2)(x – 3) = 2(x – 1)(x – 3)`
So, the value of `p(x) = 2x^2 – 8x + 6` is zero when `x – 1 = 0` or` x – 3 = 0`,
i.e., when `x = 1` or `x = 3`. So, the zeroes of `2x^2 – 8x + 6` are `1` and `3.` Observe that :
`color(blue)("Sum of its zeroes" = 1+3 =4 = (-8)/2 = ( - ( text (Coefficient of x ) ) )/( text (coefficient of) x^2 ) )`
`color(blue)("Product of its zeroes" = 1 xx 3 = 3 = 6/2 = ( text (Constant term ) )/( text (Coefficient of) x^2 ) )`
`color(red)(=>"Let us take one more quadratic polynomial")`, say, `color(blue)(p(x) = 3x^2 + 5x – 2)`
By the method of splitting the middle term,
`color(blue)(3x^2 + 5x – 2 = 3x^2 + 6x – x – 2 = 3x(x + 2) –1(x + 2))`
`" " = (3x – 1)(x + 2)`
Hence, the value of `3x^2 + 5x – 2` is zero when either `3x – 1 = 0` or` x + 2 = 0`, i.e.,
when `x = 1/3` or `x = -2` So , the zeroes of` 3x^2 + 5x – 2` are `1/3` and -2 . Observe that :
`color(blue)("Sum of its zeroes" = 1/3 + (-2) = (-5)/3 = ( ( text (Coefficient of x ) ) )/( text (Coefficient of )x^2) )`
`color(blue)("Product of its zeroes" = 1/3 xx (-2) = (-2)/3 = ( text ( Constant term ) )/( text (Coefficient of )x^2) )`
In general, if `α` and `β `are the zeroes of the quadratic polynomial `color(red)(p(x) = ax^2 + bx + c)`,
`a ≠ 0`, then you know that` x – α` and `x – β` are the factors of `p(x)`. Therefore,
`color(red)(ax^2 + bx + c = k(x – α) (x – β))`, where `k` is a constant
`" " = k [ x^2 – (α + β)x + α β]`
`" " = kx^2 – k(α + β)x + k α β`
Comparing the coefficients of `x^2, x` and constant terms on both the sides, we get
`a = k," " b = – k(α + β)` and ` " " c = kαβ`.
This gives ` color(blue)(α + β = (-b)/a)`
`color(blue)( α β = c/a)`
`α,\ \ β `are Greek letters pronounced as ‘alpha’ and ‘beta’ respectively. We will use later one more letter ‘γ’ pronounced as ‘gamma’.
i.e., `color(red)("sum of zeroes "= α + β = - b/a = ( - text ( Coefficient of x ) )/( text (Coefficient of) x^2) )`,
`color(red)("product of zeroes "= αβ = c/a = ( text (Constant term ) )/( text (Coefficient of )x^2 ) )` .
We will now try to answer the question raised regarding the relationship between zeroes
and coefficients of a quadratic polynomial.
`color(red)(=>"For this, let us take a quadratic polynomial,")` say `color(blue)(p(x) = 2x^2 – 8x + 6)`.
In Class IX, you have learnt how to factorise quadratic polynomials by splitting the middle term.
So, here we need to split the middle term ‘– 8x’ as a sum of two terms, whose product is` 6 × 2x^2 = 12x^2`.
So, we write
`color(blue)(2x^2 – 8x + 6) = 2x^2 – 6x – 2x + 6 = 2x(x – 3) – 2(x – 3)`
`" " = (2x – 2)(x – 3) = 2(x – 1)(x – 3)`
So, the value of `p(x) = 2x^2 – 8x + 6` is zero when `x – 1 = 0` or` x – 3 = 0`,
i.e., when `x = 1` or `x = 3`. So, the zeroes of `2x^2 – 8x + 6` are `1` and `3.` Observe that :
`color(blue)("Sum of its zeroes" = 1+3 =4 = (-8)/2 = ( - ( text (Coefficient of x ) ) )/( text (coefficient of) x^2 ) )`
`color(blue)("Product of its zeroes" = 1 xx 3 = 3 = 6/2 = ( text (Constant term ) )/( text (Coefficient of) x^2 ) )`
`color(red)(=>"Let us take one more quadratic polynomial")`, say, `color(blue)(p(x) = 3x^2 + 5x – 2)`
By the method of splitting the middle term,
`color(blue)(3x^2 + 5x – 2 = 3x^2 + 6x – x – 2 = 3x(x + 2) –1(x + 2))`
`" " = (3x – 1)(x + 2)`
Hence, the value of `3x^2 + 5x – 2` is zero when either `3x – 1 = 0` or` x + 2 = 0`, i.e.,
when `x = 1/3` or `x = -2` So , the zeroes of` 3x^2 + 5x – 2` are `1/3` and -2 . Observe that :
`color(blue)("Sum of its zeroes" = 1/3 + (-2) = (-5)/3 = ( ( text (Coefficient of x ) ) )/( text (Coefficient of )x^2) )`
`color(blue)("Product of its zeroes" = 1/3 xx (-2) = (-2)/3 = ( text ( Constant term ) )/( text (Coefficient of )x^2) )`
In general, if `α` and `β `are the zeroes of the quadratic polynomial `color(red)(p(x) = ax^2 + bx + c)`,
`a ≠ 0`, then you know that` x – α` and `x – β` are the factors of `p(x)`. Therefore,
`color(red)(ax^2 + bx + c = k(x – α) (x – β))`, where `k` is a constant
`" " = k [ x^2 – (α + β)x + α β]`
`" " = kx^2 – k(α + β)x + k α β`
Comparing the coefficients of `x^2, x` and constant terms on both the sides, we get
`a = k," " b = – k(α + β)` and ` " " c = kαβ`.
This gives ` color(blue)(α + β = (-b)/a)`
`color(blue)( α β = c/a)`
`α,\ \ β `are Greek letters pronounced as ‘alpha’ and ‘beta’ respectively. We will use later one more letter ‘γ’ pronounced as ‘gamma’.
i.e., `color(red)("sum of zeroes "= α + β = - b/a = ( - text ( Coefficient of x ) )/( text (Coefficient of) x^2) )`,
`color(red)("product of zeroes "= αβ = c/a = ( text (Constant term ) )/( text (Coefficient of )x^2 ) )` .