If x = \(\frac{2}{3}\) and x = – 3 are the roots of the quadratic eq

1.	If x = \(\frac{2}{3}\) and x = – 3 are the roots of the quadratic equation ax2 + 7x + b = 0, then find the values of ‘a’ and ‘b’.
Answer» Given quadratic equation is ax² + 7x + b = 0. Sum of the roots of this quadratic equation is \(\frac{-7}{a}\) . And product of roots of the quadratic equation is \(\frac{b}{a}\) Given that roots of given equation are \(\frac{2}{3}\) and – 3. Therefore, \(\frac{2}{3} + (-3) = \frac{-7}{a}\) ⇒ \(\frac{2-9}{3}\) =\(\frac{-7}{a}\) ⇒ \( \frac{-7}{3} = \frac{-7}{a}⇒a=3\) And \(\frac{2}{3}\) × (−3) = \(\frac{b}{a}⇒\frac{b}{a}\)= –2 ⇒ b = –2a = –2 × 3 = – 6. Hence, the value of and b are a = 3 and b = –6 .

If x = \(\frac{2}{3}\) and x = – 3 are the roots of the quadratic equation ax2 + 7x + b = 0, then find the values of ‘a’ and ‘b’.

Answer»

Given quadratic equation is ax² + 7x + b = 0.

Sum of the roots of this quadratic equation is \(\frac{-7}{a}\) .

And product of roots of the quadratic equation is \(\frac{b}{a}\)

Given that roots of given equation are \(\frac{2}{3}\) and – 3.

Therefore, \(\frac{2}{3} + (-3) = \frac{-7}{a}\)

⇒ \(\frac{2-9}{3}\) =\(\frac{-7}{a}\)

⇒ \( \frac{-7}{3} = \frac{-7}{a}⇒a=3\)

And \(\frac{2}{3}\) × (−3) = \(\frac{b}{a}⇒\frac{b}{a}\)= –2

⇒ b = –2a = –2 × 3 = – 6.

Hence, the value of and b are a = 3 and b = –6 .

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