Question

39.The two roots a and b of a quadratic equation are related by the equation a + b = 9 , a - b = 1 .Form an equation whose roots are 3a and b​

Answer 1

Answer

☞ Your Answer is x²-19x+60

Given

✭ Roots of a quadratic polynomial are a & b

✭ a+b = 9

✭ a-b = 1

To Find

◈ Quadratic polynomial whose roots are 3a & b

Solution

So here we are given the sum & the difference of the zeros as,

• $\tt a+b = 9 \:\:\: -eq(1)$
• $\tt a-b = 1 \:\:\: -eq(2)$

☯ $\underline{\boldsymbol{According \ to \ the \ Question}}$

$\tt \dashrightarrow a+b = 9$

$\tt \dashrightarrow a = 9-b \:\:\ -eq(3)$

Substituting the value of a in eq(2)

$\tt \twoheadrightarrow a-b = 1$

$\tt \twoheadrightarrow 9-b-b = 1$

$\tt \twoheadrightarrow 9-2b = 1$

$\tt \twoheadrightarrow -2b = 1-9$

$\tt \twoheadrightarrow -2b = -8$

$\tt \twoheadrightarrow b = \dfrac{-8}{-2}$

$\tt \red{\twoheadrightarrow b = 4}$

Substituting the value of b in eq(3)

$\tt \leadsto a = 9-b$

$\tt \leadsto a = 9-4$

$\tt \purple{\leadsto a = 5}$

So now the roots of the new polynomial will be,

$\tt :\implies 3a$

$\tt :\implies 3(5)$

$\tt \orange{:\implies a' = 15}$

Similarly,

$\tt \orange{:\implies b' = 4}$

So now we know that,

$\underline{\boxed{\tt p(x) = x^2-(a'+b')x+a'b'}}$

• a'+b' = 15+4 = 19
• a'b' = 15×4 = 60

Substituting the given values,

$\tt \dashrightarrow p(x) = x^2-(a'+b')x+a'b'$

$\tt \dashrightarrow p(x) = x^2-(19)x+60$

$\tt \pink{\dashrightarrow p(x) = x^2-19x+60}$

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Answer 2

Answer:

yess right

Step-by-step explanation:

⇢p(x)=x

2

−19x+60

this is the answer

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