To find the zeroes of the quadratic polynomial f(x) = x²-2√ax+ (a - b), we can use the quadratic formula:
x = [-b ± √(b²-4ac)] / 2a
In this case, a = 1, b = -2√a, and c = a - b. Substituting these values into the quadratic formula, we get:
x = [-(-2√a) ± √((-2√a)² - 4(1)(a-b))] / 2(1)
Simplifying the expression under the square root:
x = [2√a ± √(4a - 4a + 4b)] / 2
x = [2√a ± 2√b] / 2
x = √a ± √b
Therefore, the zeroes of the quadratic polynomial f(x) = x²-2√ax+ (a - b) are √a + √b and √a - √b.
To verify the relationship between zeroes and coefficients, we can use Vieta's formulas:
sum of zeroes = -b/a
product of zeroes = c/a
Substituting the values of a, b, and c from the given quadratic polynomial, we get:
sum of zeroes = -(-2√a) / 1 = 2√a
product of zeroes = (a-b) / 1 = a - b
We can see that the sum of the zeroes (√a + √b) + (√a - √b) = 2√a, which matches the value we obtained using Vieta's formulas. Similarly, the product of the zeroes (√a + √b) * (√a - √b) = a - b, which also matches the value we obtained using Vieta's formulas.
Therefore, we have verified the relationship between the zeroes and coefficients of the given quadratic polynomial.
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Verified answer
Answer:
To find the zeroes of the quadratic polynomial f(x) = x²-2√ax+ (a - b), we can use the quadratic formula:
x = [-b ± √(b²-4ac)] / 2a
In this case, a = 1, b = -2√a, and c = a - b. Substituting these values into the quadratic formula, we get:
x = [-(-2√a) ± √((-2√a)² - 4(1)(a-b))] / 2(1)
Simplifying the expression under the square root:
x = [2√a ± √(4a - 4a + 4b)] / 2
x = [2√a ± 2√b] / 2
x = √a ± √b
Therefore, the zeroes of the quadratic polynomial f(x) = x²-2√ax+ (a - b) are √a + √b and √a - √b.
To verify the relationship between zeroes and coefficients, we can use Vieta's formulas:
sum of zeroes = -b/a
product of zeroes = c/a
Substituting the values of a, b, and c from the given quadratic polynomial, we get:
sum of zeroes = -(-2√a) / 1 = 2√a
product of zeroes = (a-b) / 1 = a - b
We can see that the sum of the zeroes (√a + √b) + (√a - √b) = 2√a, which matches the value we obtained using Vieta's formulas. Similarly, the product of the zeroes (√a + √b) * (√a - √b) = a - b, which also matches the value we obtained using Vieta's formulas.
Therefore, we have verified the relationship between the zeroes and coefficients of the given quadratic polynomial.
[tex]\large\underline{\sf{Solution-}}[/tex]
Given polynomial is
[tex]\sf \: f(x) = {x}^{2} - 2 \sqrt{a}x + (a - b)\\ [/tex]
[tex]\sf \: f(x) = {x}^{2} - 2 \sqrt{a}x + a - b\\ [/tex]
[tex]\sf \: f(x) = [ {x}^{2} - 2 \sqrt{a}x + a] - b\\ [/tex]
[tex]\sf \: f(x) = [ {x}^{2} - 2 \sqrt{a}x + {( \sqrt{a}) }^{2} ] - b\\ [/tex]
[tex]\sf \: f(x) = {( x - \sqrt{a}) }^{2} - {( \sqrt{b}) }^{2} \\ [/tex]
[tex]\sf \: f(x) = (x - \sqrt{a} + \sqrt{b})(x - \sqrt{a} - \sqrt{b})\\ [/tex]
So, to find zeroes of f(x), we substitute
[tex]\sf \: f(x) = 0 \\ [/tex]
[tex]\sf \: (x - \sqrt{a} + \sqrt{b})(x - \sqrt{a} - \sqrt{b}) = 0\\ [/tex]
[tex]\sf \: x - \sqrt{a} + \sqrt{b} = 0 \: \: or \: \: x - \sqrt{a} - \sqrt{b} = 0\\ [/tex]
[tex]\implies\sf \: \boxed{\sf \: x = \sqrt{a} - \sqrt{b} = 0 \: \: or \: \: x = \sqrt{a} + \sqrt{b} \: }\\ [/tex]
Now, Consider
[tex]\sf \: Sum\:of\:zeroes \\ [/tex]
[tex]\sf \: = \:\sqrt{a} - \sqrt{b} + \sqrt{a} + \sqrt{b} \\ [/tex]
[tex]\sf \: = \:2\sqrt{a} \\ [/tex]
can be rewritten as
[tex]\sf \: = \: - \: \dfrac{( -2\sqrt{a} )}{1} \\ [/tex]
[tex]\sf \: = \: - \: \dfrac{\: coefficient \: of \: x \: }{coefficient \: of \: {x}^{2}} \\ [/tex]
Hence,
[tex]\sf\implies \sf \: Sum\:of\:zeroes \: = \: - \: \dfrac{\: coefficient \: of \: x \: }{coefficient \: of \: {x}^{2}} \\ \\ [/tex]
Now, Consider
[tex]\sf \: Product\:of\:zeroes \\ [/tex]
[tex]\sf \: = \: (\sqrt{a} - \sqrt{b})(\sqrt{a} + \sqrt{b}) \\ [/tex]
[tex]\sf \: = \: {( \sqrt{a}) }^{2} - {( \sqrt{b}) }^{2} \\ [/tex]
[tex]\sf \: = \: a - b \\ [/tex]
can be rewritten as
[tex]\sf \: = \: \dfrac{a - b}{1} \\ [/tex]
[tex]\sf \: = \: \dfrac{constant \: term}{coefficient \: of \: {x}^{2}} \\ \\ [/tex]
Hence,
[tex]\sf\implies \sf \:Product\:of\:zeroes \: = \: \dfrac{constant \: term}{coefficient \: of \: {x}^{2}} \\ \\ [/tex]