Question

12. In this question x^y stands for x raised to the power y. For example ,2^3=8 and 4^1.5=8. If a,b are real numbers such that a+b=3, a^2+b^2=7, the value of a^4+b^4 is? a. 49 b. 45 c. 51 d. 47 Answer: 47

Answer 1

Answer:

Option d is correct.

$a^4+b^4= 47$

Explanation:

Given: For a, b are real numbers such that a+b=3 and $a^2+b^2=7$.

Using Identities: $a^4+b^4=(a^2+b^2)^2-2a^2\cdot b^2$        ......[1]

First find the value of $a^2b^2$.

$(a+b)^2=a^2+b^2+2a\cdot b$                                  ......[2]

From the given condition, we have a+b=3 and $a^2+b^2=7$.

Substitute these values in equation [2], we have

$(3)^2=7+2ab$ or

9=7+2ab

Simplify:

ab=1 or

$a^2 \cdot b^2=1$

Now, to substitute the value of $a^2 \cdot b^2=1$ and $a^2+b^2=7$ in equation [1];

$a^4+b^4=(7)^2-2\cdot 1=49-2=47$

therefore, the value of $a^4+b^4= 47$

Answer 2

Answer:

we know the formula,

(a+b)2=a2+b2-2ab

-2ab=9-7

-2ab=2

ab=-1

By using the above formula,we get

a4+b4=(a2+b2)2-2a2b2

          =(7)2-2(-1)2

          =49-2

          =47

Step-by-step explanation: