Question

Q.1.The sum of the squares of two consecutive multiples of 7 is 637 .Write only the equation and the required multiples Q.2.Find the least positive value of p for which the given equation has equal roots . x^2+p x+4=0

Answer 1

Answer:

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Step-by-step explanation:

answer 1] Let one number be x.

Second number=x+7. (both are the multiples of 7, so they must have a difference of 7).

x²+(x+7)²=637.

x²+x²+2×x×7+49=637.

2x²+14x-588=0.

2(x²+7x-294)=0.

x²+7x-294=0/2.

x²+7x-294=0.

Now, we will solve it by Middle Term Splitting Method;

x²+21x-14x-294=0.

x(x+21)-14(x+21)=0.

(x-14)(x+21)=0.

x=+14,-21.

Hence the first number is+14 or -21.

So, the second number = 14+7 or -21+7.

= 21 or -14.

Hence the two numbers will be 14 & 21 or -14 & -21.